for(int i=1; i<100; i++) { if(________________) //填空 System.out.println(i*i/2); else System.out.println((i*i-1)/2); } 这是什么代码
时间: 2023-05-20 20:01:32 浏览: 49
这是一段使用 C 语言风格的 Java 代码。这段代码的意思是:使用一个循环,对于整数 i 从 1 到 99(不包括 100),每次加 1,如果 i 满足某个特定的条件,就打印 i 的平方除以 2 的值,否则打印 i 的平方减 1 除以 2 的值。在填空处应该填入条件表达式,即判断 i 是否满足某个条件的表达式。例如,如果想要打印 i 的平方除以 2 的值当且仅当 i 是偶数时,可以在填空处填入 i % 2 == 0。
相关问题
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int MAXN = 107; const int INF = 0x3f3f3f3f; typedef long long ll; int graph[MAXN][MAXN]; int main() { int n, e; while (~scanf("%d %d", &n, &e)) { memset(graph, INF, sizeof(graph)); for (int i = 1; i <= e; i++) { int from, to, v; scanf("%d %d %d", &from, &to, &v); ++from, ++to; if (from == to) { graph[from][to] = 0; } else { graph[from][to] = v; graph[to][from] = v; } } for (int k = 1; k <= n; k++) { for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (graph[i][j] > graph[i][k] + graph[k][j]) { graph[i][j] = graph[i][k] + graph[k][j]; } } } } ll dis[MAXN] = {0}; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (i != j) { dis[i] += (ll)graph[i][j]; } } } ll min_dis = INF; int min_pos = 1; for (int i = 1; i <= n; i++) { if (dis[i] < min_dis) { min_dis = dis[i]; min_pos = i; } } --min_pos; printf("%d\n", min_pos); } return 0; }帮我把这段代码转换为c++
在这里提供一份已经是 C++ 代码的版本,可以直接使用:
```cpp
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 107;
const int INF = 0x3f3f3f3f;
typedef long long ll;
int graph[MAXN][MAXN];
int main() {
int n, e;
while (cin >> n >> e) {
memset(graph, INF, sizeof(graph));
for (int i = 1; i <= e; i++) {
int from, to, v;
cin >> from >> to >> v;
++from, ++to;
if (from == to) {
graph[from][to] = 0;
} else {
graph[from][to] = v;
graph[to][from] = v;
}
}
for (int k = 1; k <= n; k++) {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (graph[i][j] > graph[i][k] + graph[k][j]) {
graph[i][j] = graph[i][k] + graph[k][j];
}
}
}
}
ll dis[MAXN] = {0};
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (i != j) {
dis[i] += (ll)graph[i][j];
}
}
}
ll min_dis = INF;
int min_pos = 1;
for (int i = 1; i <= n; i++) {
if (dis[i] < min_dis) {
min_dis = dis[i];
min_pos = i;
}
}
--min_pos;
cout << min_pos << endl;
}
return 0;
}
```
#define _CRT_SECURE_NO_WARNINGS 1 #include<stdio.h> int main() { int s = 0, a[1001] = {0}; for (int i = 1; i <= 500; i++) { int t = 0; for (int j = 1; j <= i; j++) { if (i%j == 0) { a[t++] = j; } } int p=0; for (int k = 0; k < t; k++) { p += a[k]; } if (p == i) s += i; } printf("%d\n", s); return 0; }改错
在程序中,第一层循环应该是 `i <= 1000` 而不是 `i <= 500`,因为题目要求的是 1000 以内的完全数,而不是 500 以内的完全数。此外,在程序中 `a[1001]` 的长度应该是 `a[1000]`,因为数组下标是从 0 开始的。修改后的程序如下:
```c
#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>
int main()
{
int s = 0, a[1000] = {0};
for (int i = 1; i <= 1000; i++)
{
int t = 0;
for (int j = 1; j <= i; j++)
{
if (i % j == 0)
{
a[t++] = j;
}
}
int p = 0;
for (int k = 0; k < t; k++)
{
p += a[k];
}
if (p == i)
s += i;
}
printf("%d\n", s);
return 0;
}
```
相关推荐
![c](https://img-home.csdnimg.cn/images/20210720083646.png)
![pdf](https://img-home.csdnimg.cn/images/20210720083512.png)
![application/x-rar](https://img-home.csdnimg.cn/images/20210720083606.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)