for (int i = 1,s = o; i <= n; i++) { int t = 1; for (int j = 1;j <= i; j++) t *= j; s += t;}
时间: 2023-12-06 22:03:30 浏览: 38
This code is incomplete as the increment statements for both the outer and inner loops are missing. The code should be modified as follows:
```
for (int i = 1, s = o; i <= n; i++) {
int t = 1;
for (int j = 1; j <= i; j++)
t *= j;
s += t;
}
```
This code calculates the sum of factorials from 1 to n and stores it in the variable s. The outer loop iterates from 1 to n, and the inner loop calculates the factorial of each number using the formula t *= j. The result is added to the variable s in each iteration of the outer loop. The initial value of s is stored in the variable o.
相关问题
int s = 0; for (int i = 1;i < n – 1;i++) for (int j = i + 1;j <= n;j++) s++;的时间复杂度
这段代码中,第一个for循环的循环条件是`i < n - 1`,而每次循环`i`的增量是`i++`,因此第一个for循环的时间复杂度为O(n)。
第二个for循环的循环条件是`j <= n`,而每次循环`j`的增量是`j++`,因此第二个for循环的时间复杂度也为O(n)。
两个for循环的嵌套使得内层for循环的循环体被执行了`(n-2) * (n-1)`次,因此总的时间复杂度为O(n^2)。
最后,代码中`s`的语句没有实际意义,可能是一个错误。
#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;
}
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