STDIN → STDOUT
ACM 模式输入输出模板
C++、Python、Java、Go 的 ACM 模式（标准输入输出）读写模板：单组、读到 EOF、T 组数据、n 加数组、矩阵、字符网格、建图、整行读入和快读快写，每种都附说明和可运行样例。
Stdin/stdout templates for ACM-style problems and OAs in C++, Python, Java and Go: EOF loops, T test cases, arrays, matrices, grids, graphs, whole lines and fast I/O, each with a runnable sample.
单组输入Single test case
只有一组数据，读完直接输出。按「空白分隔的记号」读（split / cin >> / Fscan / Java 的 next()），换行和多余空格都不影响。示例：输出 a + b。
One test case: read it, print the answer. Reading whitespace-separated tokens (split / cin >> / Fscan / Java's next()) ignores line breaks and extra spaces. Example: print a + b.
3 5
8
C++
#include <iostream>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
long long a, b;
cin >> a >> b;
cout << a + b << '\n';
return 0;
}
Python 3
import sys
def main():
data = sys.stdin.read().split()
a, b = int(data[0]), int(data[1])
print(a + b)
main()
Java
import java.io.*;
import java.util.*;
public class Main {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st = new StringTokenizer("");
// Next whitespace-separated token; line breaks don't matter.
static String next() throws IOException {
while (!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public static void main(String[] args) throws IOException {
long a = Long.parseLong(next());
long b = Long.parseLong(next());
System.out.println(a + b);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
)
func main() {
in := bufio.NewReader(os.Stdin)
var a, b int64
fmt.Fscan(in, &a, &b)
fmt.Println(a + b)
}
多组输入，读到文件结束（EOF）Read until end of input (EOF)
题目不告诉组数时，一直读到输入结束。C++ 用 while (cin >> a >> b)，Java 判断 readLine() 返回 null，Go 判断 Fscan 的 err，Python 直接 for line in sys.stdin。
When the number of cases isn't given, read until input runs out: while (cin >> a >> b) in C++, readLine() == null in Java, Fscan's err in Go, and for line in sys.stdin in Python.
1 2 3 4 10 20
3 7 30
C++
#include <iostream>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
long long a, b;
while (cin >> a >> b) {
cout << a + b << '\n';
}
return 0;
}
Python 3
import sys
def main():
out = []
for line in sys.stdin:
parts = line.split()
if not parts:
continue
a, b = int(parts[0]), int(parts[1])
out.append(str(a + b))
print('\n'.join(out))
main()
Java
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder out = new StringBuilder();
String line;
while ((line = br.readLine()) != null) {
StringTokenizer st = new StringTokenizer(line);
if (!st.hasMoreTokens()) continue;
long a = Long.parseLong(st.nextToken());
long b = Long.parseLong(st.nextToken());
out.append(a + b).append('\n');
}
System.out.print(out);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
)
func main() {
in := bufio.NewReader(os.Stdin)
out := bufio.NewWriter(os.Stdout)
defer out.Flush()
var a, b int64
for {
if _, err := fmt.Fscan(in, &a, &b); err != nil {
break
}
fmt.Fprintln(out, a+b)
}
}
首行给出组数 TT test cases on the first line
第一行是组数 T，后面每组格式相同。循环 T 次，每组答案单独一行；输出先攒起来最后一次打印，比每组 print 快。
Line 1 holds T, the number of cases; each case has the same format. Loop T times and print one answer per line. Collecting output and printing once is faster than printing per case.
3 1 2 3 4 5 6
3 7 11
C++
#include <iostream>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) {
long long a, b;
cin >> a >> b;
cout << a + b << '\n';
}
return 0;
}
Python 3
import sys
def main():
input = sys.stdin.readline
t = int(input())
out = []
for _ in range(t):
a, b = map(int, input().split())
out.append(str(a + b))
print('\n'.join(out))
main()
Java
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());
StringBuilder out = new StringBuilder();
while (t-- > 0) {
StringTokenizer st = new StringTokenizer(br.readLine());
long a = Long.parseLong(st.nextToken());
long b = Long.parseLong(st.nextToken());
out.append(a + b).append('\n');
}
System.out.print(out);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
)
func main() {
in := bufio.NewReader(os.Stdin)
out := bufio.NewWriter(os.Stdout)
defer out.Flush()
var t int
fmt.Fscan(in, &t)
for ; t > 0; t-- {
var a, b int64
fmt.Fscan(in, &a, &b)
fmt.Fprintln(out, a+b)
}
}
多组输入，以 0 0 结束Stop at a terminator (0 0)
读到约定的结束标记（如 0 0）就停，标记本身不处理、不输出。先判断标记再计算。
Stop at the agreed terminator, such as 0 0. The terminator itself is neither processed nor printed, so check it before computing.
1 2 3 4 0 0
3 7
C++
#include <iostream>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
long long a, b;
while (cin >> a >> b && !(a == 0 && b == 0)) {
cout << a + b << '\n';
}
return 0;
}
Python 3
import sys
def main():
out = []
for line in sys.stdin:
parts = line.split()
if not parts:
continue
a, b = int(parts[0]), int(parts[1])
if a == 0 and b == 0:
break
out.append(str(a + b))
print('\n'.join(out))
main()
Java
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder out = new StringBuilder();
String line;
while ((line = br.readLine()) != null) {
StringTokenizer st = new StringTokenizer(line);
if (!st.hasMoreTokens()) continue;
long a = Long.parseLong(st.nextToken());
long b = Long.parseLong(st.nextToken());
if (a == 0 && b == 0) break;
out.append(a + b).append('\n');
}
System.out.print(out);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
)
func main() {
in := bufio.NewReader(os.Stdin)
out := bufio.NewWriter(os.Stdout)
defer out.Flush()
var a, b int64
for {
if _, err := fmt.Fscan(in, &a, &b); err != nil || (a == 0 && b == 0) {
break
}
fmt.Fprintln(out, a+b)
}
}
先给 n，再给一行 n 个数n, then an array of n numbers
最常见的形态：第一行 n，第二行 n 个整数。求和可能超过 int 范围，用 long long / long / int64。示例：输出数组之和。
The most common shape: n on line 1, then n integers. A sum can overflow int, so use long long / long / int64. Example: print the sum.
5 3 1 4 1 5
14
C++
#include <iostream>
#include <vector>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<long long> nums(n);
for (auto &x : nums) cin >> x;
long long sum = 0;
for (long long x : nums) sum += x;
cout << sum << '\n';
return 0;
}
Python 3
import sys
def main():
data = sys.stdin.read().split()
n = int(data[0])
nums = list(map(int, data[1:1 + n]))
print(sum(nums))
main()
Java
import java.io.*;
import java.util.*;
public class Main {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st = new StringTokenizer("");
// Next whitespace-separated token; line breaks don't matter.
static String next() throws IOException {
while (!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public static void main(String[] args) throws IOException {
int n = Integer.parseInt(next());
long[] nums = new long[n];
for (int i = 0; i < n; i++) nums[i] = Long.parseLong(next());
long sum = 0;
for (long x : nums) sum += x;
System.out.println(sum);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
)
func main() {
in := bufio.NewReader(os.Stdin)
var n int
fmt.Fscan(in, &n)
nums := make([]int64, n)
for i := range nums {
fmt.Fscan(in, &nums[i])
}
var sum int64
for _, x := range nums {
sum += x
}
fmt.Println(sum)
}
n × m 矩阵n × m matrix
第一行 n m，接着 n 行、每行 m 个数。用二维数组存，下标从 0 开始。示例：输出每一行的和。
Line 1 is n m, then n rows of m numbers each. Store a 2D array indexed from 0. Example: print the sum of each row.
2 3 1 2 3 4 5 6
6 15
C++
#include <iostream>
#include <vector>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
cin >> n >> m;
vector<vector<long long>> grid(n, vector<long long>(m));
for (auto &row : grid)
for (auto &x : row) cin >> x;
for (const auto &row : grid) {
long long sum = 0;
for (long long x : row) sum += x;
cout << sum << '\n';
}
return 0;
}
Python 3
import sys
def main():
input = sys.stdin.readline
n, m = map(int, input().split())
grid = [list(map(int, input().split()))[:m] for _ in range(n)]
print('\n'.join(str(sum(row)) for row in grid))
main()
Java
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int m = Integer.parseInt(st.nextToken());
long[][] grid = new long[n][m];
for (int i = 0; i < n; i++) {
st = new StringTokenizer(br.readLine());
for (int j = 0; j < m; j++) grid[i][j] = Long.parseLong(st.nextToken());
}
StringBuilder out = new StringBuilder();
for (long[] row : grid) {
long sum = 0;
for (long x : row) sum += x;
out.append(sum).append('\n');
}
System.out.print(out);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
)
func main() {
in := bufio.NewReader(os.Stdin)
out := bufio.NewWriter(os.Stdout)
defer out.Flush()
var n, m int
fmt.Fscan(in, &n, &m)
grid := make([][]int64, n)
for i := range grid {
grid[i] = make([]int64, m)
for j := range grid[i] {
fmt.Fscan(in, &grid[i][j])
}
}
for _, row := range grid {
var sum int64
for _, x := range row {
sum += x
}
fmt.Fprintln(out, sum)
}
}
字符网格（迷宫 / 地图）Character grid (maze / map)
第一行 n m，接着 n 行字符串，每行 m 个字符，常见于迷宫和 BFS 题。按字符串整行读，再用 grid[i][j] 取格子。示例：数出 # 的个数。
Line 1 is n m, then n strings of m characters, typical for mazes and BFS. Read each row as a string and index grid[i][j]. Example: count the # cells.
3 4 #..# .##. ....
4
C++
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
cin >> n >> m;
vector<string> grid(n);
for (auto &row : grid) cin >> row;
int count = 0;
for (const auto &row : grid)
for (char c : row) count += c == '#';
cout << count << '\n';
return 0;
}
Python 3
import sys
def main():
input = sys.stdin.readline
n, m = map(int, input().split())
grid = [input().strip() for _ in range(n)]
print(sum(row.count('#') for row in grid))
main()
Java
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int m = Integer.parseInt(st.nextToken());
char[][] grid = new char[n][];
for (int i = 0; i < n; i++) grid[i] = br.readLine().trim().toCharArray();
int count = 0;
for (char[] row : grid)
for (char c : row) if (c == '#') count++;
System.out.println(count);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
)
func main() {
in := bufio.NewReader(os.Stdin)
var n, m int
fmt.Fscan(in, &n, &m)
grid := make([]string, n)
for i := range grid {
fmt.Fscan(in, &grid[i])
}
count := 0
for _, row := range grid {
for _, c := range row {
if c == '#' {
count++
}
}
}
fmt.Println(count)
}
n 个点、m 条边建图Graph from n nodes and m edges
第一行 n m，接着 m 行 u v（点编号从 1 开始）。用邻接表存图，无向图两个方向都要加。示例：从 1 号点 BFS，输出到每个点的最短距离，走不到输出 -1。
Line 1 is n m, then m lines u v with nodes numbered from 1. Store an adjacency list and add both directions for an undirected graph. Example: BFS from node 1 and print each distance, or -1 if unreachable.
5 3 1 2 2 3 1 4
0 1 2 1 -1
C++
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
cin >> n >> m;
vector<vector<int>> adj(n + 1);
for (int i = 0; i < m; i++) {
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
vector<int> dist(n + 1, -1);
queue<int> q;
dist[1] = 0;
q.push(1);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int v : adj[u]) {
if (dist[v] == -1) {
dist[v] = dist[u] + 1;
q.push(v);
}
}
}
for (int i = 1; i <= n; i++) cout << dist[i] << (i == n ? '\n' : ' ');
return 0;
}
Python 3
import sys
from collections import deque
def main():
it = iter(map(int, sys.stdin.read().split()))
n, m = next(it), next(it)
adj = [[] for _ in range(n + 1)]
for _ in range(m):
u, v = next(it), next(it)
adj[u].append(v)
adj[v].append(u)
dist = [-1] * (n + 1)
dist[1] = 0
queue = deque([1])
while queue:
u = queue.popleft()
for v in adj[u]:
if dist[v] == -1:
dist[v] = dist[u] + 1
queue.append(v)
print(' '.join(map(str, dist[1:])))
main()
Java
import java.io.*;
import java.util.*;
public class Main {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st = new StringTokenizer("");
// Next whitespace-separated token; line breaks don't matter.
static String next() throws IOException {
while (!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public static void main(String[] args) throws IOException {
int n = Integer.parseInt(next());
int m = Integer.parseInt(next());
List<List<Integer>> adj = new ArrayList<>();
for (int i = 0; i <= n; i++) adj.add(new ArrayList<>());
for (int i = 0; i < m; i++) {
int u = Integer.parseInt(next());
int v = Integer.parseInt(next());
adj.get(u).add(v);
adj.get(v).add(u);
}
int[] dist = new int[n + 1];
Arrays.fill(dist, -1);
ArrayDeque<Integer> queue = new ArrayDeque<>();
dist[1] = 0;
queue.add(1);
while (!queue.isEmpty()) {
int u = queue.poll();
for (int v : adj.get(u)) {
if (dist[v] == -1) {
dist[v] = dist[u] + 1;
queue.add(v);
}
}
}
StringBuilder out = new StringBuilder();
for (int i = 1; i <= n; i++) out.append(dist[i]).append(i == n ? '\n' : ' ');
System.out.print(out);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
)
func main() {
in := bufio.NewReader(os.Stdin)
out := bufio.NewWriter(os.Stdout)
defer out.Flush()
var n, m int
fmt.Fscan(in, &n, &m)
adj := make([][]int, n+1)
for i := 0; i < m; i++ {
var u, v int
fmt.Fscan(in, &u, &v)
adj[u] = append(adj[u], v)
adj[v] = append(adj[v], u)
}
dist := make([]int, n+1)
for i := range dist {
dist[i] = -1
}
dist[1] = 0
queue := []int{1}
for len(queue) > 0 {
u := queue[0]
queue = queue[1:]
for _, v := range adj[u] {
if dist[v] == -1 {
dist[v] = dist[u] + 1
queue = append(queue, v)
}
}
}
for i := 1; i <= n; i++ {
if i > 1 {
out.WriteByte(' ')
}
fmt.Fprint(out, dist[i])
}
out.WriteByte('\n')
}
整行读入（含空格的字符串）Whole lines with spaces
字符串里有空格时要按整行读。坑：C++ 先 cin >> n 再 getline，会读到 n 后面剩下的空行，要先多 getline 一次；Go 同理。示例：输出每行的单词数。
Read whole lines when strings contain spaces. Pitfall: after cin >> n, the next getline returns the rest of n's line, so call getline once more first (same in Go). Example: print the word count of each line.
3 hello world a b c single
2 3 1
C++
#include <iostream>
#include <sstream>
#include <string>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
string line;
getline(cin, line); // the rest of n's line
for (int i = 0; i < n; i++) {
getline(cin, line);
istringstream words(line);
string word;
int count = 0;
while (words >> word) count++;
cout << count << '\n';
}
return 0;
}
Python 3
import sys
def main():
input = sys.stdin.readline
n = int(input())
out = []
for _ in range(n):
line = input().rstrip('\r\n')
out.append(str(len(line.split())))
print('\n'.join(out))
main()
Java
import java.io.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine().trim());
StringBuilder out = new StringBuilder();
for (int i = 0; i < n; i++) {
String line = br.readLine().trim();
out.append(line.isEmpty() ? 0 : line.split("\\s+").length).append('\n');
}
System.out.print(out);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
"strings"
)
func main() {
in := bufio.NewReader(os.Stdin)
out := bufio.NewWriter(os.Stdout)
defer out.Flush()
var n int
fmt.Fscan(in, &n)
in.ReadString('\n') // the rest of n's line
for i := 0; i < n; i++ {
line, _ := in.ReadString('\n')
fmt.Fprintln(out, len(strings.Fields(line)))
}
}
每行个数不定，读到结束Lines of varying length until EOF
每行数字个数不固定、行数也不给时，逐行读，再把这一行拆开。空行直接跳过。示例：输出每行之和。
When neither the count per line nor the number of lines is given, read line by line and split each one. Blank lines are skipped. Example: print each line's sum.
1 2 3 4 5 6
6 4 11
C++
#include <iostream>
#include <sstream>
#include <string>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
string line;
while (getline(cin, line)) {
istringstream nums(line);
long long x, sum = 0;
bool any = false;
while (nums >> x) {
sum += x;
any = true;
}
if (any) cout << sum << '\n';
}
return 0;
}
Python 3
import sys
def main():
out = []
for line in sys.stdin:
if line.strip():
out.append(str(sum(map(int, line.split()))))
print('\n'.join(out))
main()
Java
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder out = new StringBuilder();
String line;
while ((line = br.readLine()) != null) {
StringTokenizer st = new StringTokenizer(line);
if (!st.hasMoreTokens()) continue;
long sum = 0;
while (st.hasMoreTokens()) sum += Long.parseLong(st.nextToken());
out.append(sum).append('\n');
}
System.out.print(out);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
"strconv"
"strings"
)
func main() {
sc := bufio.NewScanner(os.Stdin)
sc.Buffer(make([]byte, 1<<20), 1<<20) // lines up to 1 MB
out := bufio.NewWriter(os.Stdout)
defer out.Flush()
for sc.Scan() {
fields := strings.Fields(sc.Text())
if len(fields) == 0 {
continue
}
var sum int64
for _, f := range fields {
x, _ := strconv.ParseInt(f, 10, 64)
sum += x
}
fmt.Fprintln(out, sum)
}
}
空格分隔输出（行末无空格）Space-separated output, no trailing space
一行输出多个数时用空格隔开。多数评测会忽略行末空格，但有的严格比对；用 join 或「不是第一个才加空格」最稳。示例：把数组倒序输出。
Separate the numbers on a line with spaces. Most judges ignore trailing spaces but some compare exactly; join, or add a space before every item except the first. Example: print the array reversed.
4 1 2 3 4
4 3 2 1
C++
#include <iostream>
#include <vector>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<long long> nums(n);
for (auto &x : nums) cin >> x;
for (int i = n - 1; i >= 0; i--) {
cout << nums[i] << (i == 0 ? '\n' : ' ');
}
return 0;
}
Python 3
import sys
def main():
data = sys.stdin.read().split()
n = int(data[0])
nums = data[1:1 + n]
print(' '.join(reversed(nums)))
main()
Java
import java.io.*;
import java.util.*;
public class Main {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st = new StringTokenizer("");
// Next whitespace-separated token; line breaks don't matter.
static String next() throws IOException {
while (!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public static void main(String[] args) throws IOException {
int n = Integer.parseInt(next());
String[] nums = new String[n];
for (int i = 0; i < n; i++) nums[i] = next();
StringJoiner out = new StringJoiner(" ");
for (int i = n - 1; i >= 0; i--) out.add(nums[i]);
System.out.println(out);
}
}
Go
package main
import (
"bufio"
"fmt"
"os"
"strings"
)
func main() {
in := bufio.NewReader(os.Stdin)
var n int
fmt.Fscan(in, &n)
nums := make([]string, n)
for i := range nums {
fmt.Fscan(in, &nums[i])
}
for i, j := 0, n-1; i < j; i, j = i+1, j-1 {
nums[i], nums[j] = nums[j], nums[i]
}
fmt.Println(strings.Join(nums, " "))
}
快读快写（大数据量）Fast I/O for large input
输入输出到 10^5～10^6 量级时，I/O 本身就可能超时。C++ 关同步（sync_with_stdio(false) + cin.tie(nullptr)）并用 \n 代替 endl；Java 用 BufferedReader + StreamTokenizer 读、PrintWriter 写（StreamTokenizer 把数读成 double，超过 2^53 的整数会失真）；Go 用 bufio.Scanner 按词读、bufio.Writer 写；Python 用 sys.stdin.buffer 一次读完、最后一次 write。示例：输出前缀和。
At 10^5 to 10^6 numbers, I/O alone can time out. C++: turn off sync (sync_with_stdio(false) + cin.tie(nullptr)) and use \n instead of endl. Java: BufferedReader + StreamTokenizer to read, PrintWriter to write (StreamTokenizer reads doubles, so integers past 2^53 lose precision). Go: bufio.Scanner by words, bufio.Writer for output. Python: read sys.stdin.buffer at once and write once. Example: print prefix sums.
5 1 2 3 4 5
1 3 6 10 15
C++
#include <iostream>
#include <string>
using namespace std;
int main() {
ios::sync_with_stdio(false); // stop syncing with C stdio
cin.tie(nullptr); // don't flush cout before each read
int n;
cin >> n;
long long total = 0;
string out;
for (int i = 0; i < n; i++) {
long long x;
cin >> x;
total += x;
out += to_string(total);
out += '\n'; // endl would flush every line
}
cout << out;
return 0;
}
Python 3
import sys
def main():
data = sys.stdin.buffer.read().split()
n = int(data[0])
out = []
total = 0
for i in range(1, n + 1):
total += int(data[i])
out.append(total)
sys.stdout.write('\n'.join(map(str, out)) + '\n')
main()
Java
import java.io.*;
public class Main {
public static void main(String[] args) throws IOException {
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
in.nextToken();
int n = (int) in.nval;
long total = 0;
for (int i = 0; i < n; i++) {
in.nextToken();
total += (long) in.nval; // a double: exact up to 2^53
out.println(total);
}
out.flush();
}
}
Go
package main
import (
"bufio"
"os"
"strconv"
)
func main() {
sc := bufio.NewScanner(os.Stdin)
sc.Buffer(make([]byte, 1<<20), 1<<20)
sc.Split(bufio.ScanWords)
readInt := func() int64 {
sc.Scan()
x, _ := strconv.ParseInt(sc.Text(), 10, 64)
return x
}
out := bufio.NewWriter(os.Stdout)
defer out.Flush()
n := int(readInt())
var total int64
for i := 0; i < n; i++ {
total += readInt()
out.WriteString(strconv.FormatInt(total, 10))
out.WriteByte('\n')
}
}
写题时在编辑器工具栏切到 ACM 模式，选「输入模板」即可换成其中任一份完整程序（改过的代码会先确认）。 In the practice editor, switch to ACM mode and pick Input templates to start from any of these whole programs (edited code asks first). 去题库练习 · Open the problems →
语言版本、编译命令、时间内存和递归深度限制见 Versions, compile commands and limits: 判题环境说明 · Judge environment →