for(int i=0,j=10; i=j=10; i++,j--)
时间: 2023-12-11 17:02:11 浏览: 26
This code snippet is not valid as the conditional statement in the for loop is not properly defined. The statement "i=j=10" is not a valid comparison operator.
A valid for loop statement would have a comparison operator in the conditional statement, such as:
for(int i=0,j=10; i<10 && j>0; i++,j--) {
// loop code here
}
This loop will run as long as i is less than 10 AND j is greater than 0. The loop code will execute each time the loop iterates.
相关问题
for (int i = 0; i < 5; i ++ ) for (int j = i + 1; j <= 10 - i; j ++ )
这是一个双重循环语句,其中包含两个 for 循环。第一个循环变量 i 的初始值为 0,每次循环增加 1,当 i 的值为 5 时结束循环。第二个循环变量 j 的初始值为 i+1,每次循环增加 1,当 j 的值大于等于 10-i 时结束循环。这段代码的作用是对一些数据进行遍历,具体操作需要看循环体内部的代码。
#include<iostream> #include<string> #include<math.h> using namespace std; int main(){ int m,n; cin>>m>>n; int a[m+5][n+5]; for(int i=1;i<=m;i++) for(int j=1;j<=n;j++) cin>>a[i][j]; int r,s; cin>>r>>s; int b[r+5][s+5]; for(int i=1;i<=r;i++) for(int j=1;j<=s;j++) cin>>b[i][j]; int minX=1; int minY=1; int min=10000000; for(int u=1;u<=m-r+1;u++) for(int v=1;v<=n-s+1;v++){ int sum=0; for(int i=1;i<=r;i++){ for(int j=1;j<=s;j++){ sum+=abs(a[u+i-1][v+j-1]-b[i][j]); } } if(sum<min){ minX=u; minY=v; min=sum; } } for(int i=1;i<=r;i++){ for(int j=1;j<=s;j++){ cout<<a[minX+i-1][minY+j-1]<<' '; } cout<<endl; } return 0; }改为c语言
把代码中的"cout"改为"printf",去掉命名空间,整体代码如下:
#include <stdio.h>
#include <string.h>
#include <math.h>
int main() {
int m, n;
scanf("%d%d", &m, &n);
int a[m + 5][n + 5];
for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
scanf("%d", &a[i][j]);
int r, s;
scanf("%d%d", &r, &s);
int b[r + 5][s + 5];
for (int i = 1; i <= r; i++)
for (int j = 1; j <= s; j++)
scanf("%d", &b[i][j]);
int minX = 1;
int minY = 1;
int min = 10000000;
for (int u = 1; u <= m - r + 1; u++)
for (int v = 1; v <= n - s + 1; v++) {
int sum = 0;
for (int i = 1; i <= r; i++) {
for (int j = 1; j <= s; j++) {
sum += abs(a[u + i - 1][v + j - 1] - b[i][j]);
}
}
if (sum < min) {
minX = u;
minY = v;
min = sum;
}
}
for (int i = 1; i <= r; i++) {
for (int j = 1; j <= s; j++) {
printf("%d ", a[minX + i - 1][minY + j - 1]);
}
printf("\n");
}
return 0;
}