for (int i = 1; i <= 9; i++) { for (int j = 1; j <= 1; j++) { System.out.print(j + "*" + i + "=" + (i * j) + "\t"); } System.out.println(); } 结果怎么改为9*9的乘法表
时间: 2024-03-21 07:39:34 浏览: 23
要将这段代码修改为打印 9*9 的乘法表,需要将内层循环的条件改为 j <= i。同时,外层循环的终止条件也应该是 i <= 9。修改后的代码如下:
```java
for (int i = 1; i <= 9; i++) {
for (int j = 1; j <= i; j++) {
System.out.print(j + "*" + i + "=" + (i * j) + "\t");
}
System.out.println();
}
```
这样就可以打印出完整的 9*9 的乘法表了。
相关问题
#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;
}
帮我优化下面的代码:qint32 press = channel/2048; QVector<double> temp(channel); if (press == 1){ for (int j = 0;j<16;j++){ for(int i = 0;i<128;i++){ temp<< receiveData[i+j136+2].toInt(nullptr,16); }} }else if (press > 1){ for (int j = 0;j<16;j++){ for(int i = 0;i<128;i++){ for (int k = 0;k<press;k++){ temp<<receiveData[i+j136+2].toInt(nullptr,16); } }} }else if (press < 1){ for (int j = 0;j<16;j++){ for(int i = 0;i<128press;i++){ double s = 0; for (int k = 0;k<int(1/press);k++){ s += receiveData[i+j136+2+k].toInt(nullptr,16); } temp<<s; }} } for (int i = 0 ;i<channel;i++) { scatterEX[i+firstchannel] += temp[i]; }
可以尝试将三个if语句合并为一个,避免重复的循环。同时,可以使用预处理器指令定义常量,以避免硬编码。最后,可以考虑使用并行化的方法来加速循环计算。优化后的代码如下:
```
#define BLOCK_SIZE 128
#define HEX_BASE 16
qint32 press = channel / 2048;
QVector<double> temp(channel);
#pragma omp parallel for
for (int j = 0; j < 16; j++) {
for (int i = 0; i < BLOCK_SIZE; i++) {
int index = i + j * BLOCK_SIZE + 2;
if (press == 1) {
temp << receiveData[index].toInt(nullptr, HEX_BASE);
} else if (press > 1) {
for (int k = 0; k < press; k++) {
temp << receiveData[index].toInt(nullptr, HEX_BASE);
}
} else if (press < 1) {
double s = 0;
for (int k = 0; k < int(1 / press); k++) {
s += receiveData[index + k].toInt(nullptr, HEX_BASE);
}
temp << s;
}
}
}
#pragma omp parallel for
for (int i = 0; i < channel; i++) {
scatterEX[i + firstchannel] += temp[i];
}
```
注意:如果原代码中的变量 `firstchannel` 是在其他地方定义的,需要在优化后的代码中进行相应的修改。