C语言编程：不用函数，用三种循环结构计算1x1+2x2+3x3+...+100x100的值？ 用c语言编写1+2+3+...+100
C\u8bed\u8a00\u5faa\u73af\u9898\uff01\u7f16\u5199\u4e00\u4e2a\u7a0b\u5e8f\uff0c\u6c42s=1+(1+2)+(1+2+3)+...+(1+2+3+...+n)\u3002\u4fee\u6539\u5982\u4e0b
#include
int main(void)
{
int i,j,n;
int item,s; //\u6b64\u5904\u6570\u636e\u7c7b\u578b\u5b9a\u4e49\u9519\u8bef\uff1b
scanf("%d",&n);
s=0; n=5;
for(i=1;i<=n;i++){
item=0; //item\u5e94\u8be5\u662f\u4ece0\u5f00\u59cb\uff1b
for(j=1;j<=i;j++) //\u7b2c\u4e8c\u5c42\u5faa\u73af\u5e94\u8be5\u5c0f\u4e8e\u7b49\u4e8ei\uff1b
item=item+j;
s=s+item;
}
printf("%d\n",s);
return 0;
}
\u56fe\u4e3a\u539f\u7a0b\u5e8f\u5728\u7b2c\u4e8c\u5c42\u5faa\u73af\u52a0\u4e86\u4e2a\u8f93\u51fa\uff0c\u67e5\u51fa\u7b2c\u4e8c\u5c42\u5faa\u73af\u6709\u95ee\u9898\u3002
\u6269\u5c55\u8d44\u6599\uff1a
c\u8bed\u8a00\u4e66\u5199\u89c4\u5219\uff1a
1.\u4e00\u4e2a\u8bf4\u660e\u6216\u4e00\u4e2a\u8bed\u53e5\u5360\u4e00\u884c\u3002
2.\u7528{} \u62ec\u8d77\u6765\u7684\u90e8\u5206\uff0c\u901a\u5e38\u8868\u793a\u4e86\u7a0b\u5e8f\u7684\u67d0\u4e00\u5c42\u6b21\u7ed3\u6784\u3002{}\u4e00\u822c\u4e0e\u8be5\u7ed3\u6784\u8bed\u53e5\u7684\u7b2c\u4e00\u4e2a\u5b57\u6bcd\u5bf9\u9f50\uff0c\u5e76\u5355\u72ec\u5360\u4e00\u884c\u3002
3.\u4f4e\u4e00\u5c42\u6b21\u7684\u8bed\u53e5\u6216\u8bf4\u660e\u53ef\u6bd4\u9ad8\u4e00\u5c42\u6b21\u7684\u8bed\u53e5\u6216\u8bf4\u660e\u7f29\u8fdb\u82e5\u5e72\u683c\u540e\u4e66\u5199\u3002\u4ee5\u4fbf\u770b\u8d77\u6765\u66f4\u52a0\u6e05\u6670\uff0c\u589e\u52a0\u7a0b\u5e8f\u7684\u53ef\u8bfb\u6027\u3002\u5728\u7f16\u7a0b\u65f6\u5e94\u529b\u6c42\u9075\u5faa\u8fd9\u4e9b\u89c4\u5219\uff0c\u4ee5\u517b\u6210\u826f\u597d\u7684\u7f16\u7a0b\u98ce\u683c\u3002
\u53c2\u8003\u8d44\u6599\uff1a
\u767e\u5ea6\u767e\u79d1-c\u8bed\u8a00
#include
#include
int main(){
int sum=0,i=1;
while(i<=100){
sum=sum+i;
i++;
}
printf("1+2+3+...+100=%d",sum);
getch();
}
\u65b9\u6cd5\u4e8c\u3001
int main(){
int sum=0,i;
for(i=1;i<=100;i++){
sum=sum+i;
}
printf("1+2+3+...+100=%d",sum);
getch();
}
\u65b9\u6cd5\u4e09\u3001
int main(){
int sum=0,i=1;
do{
sum=sum+i;
i++;
}
while(i<=100);
printf("1+2+3+...+100=%d",sum);
getch();
}
\u6269\u5c55\u8d44\u6599\uff1ajava\u4e2d\u7f16\u7a0b\u5b9e\u73b01+2+3+4...+100\u7684\u548c\u3002
\u5b9a\u4e49\u4e00\u4e2a\u53d8\u91cftotal\uff0c\u7528\u6765\u7edf\u8ba1\u548c\uff1b\u7136\u540e\u7528\u5faa\u73af\uff0c\u4ece1\u5230100\uff1b\u6bcf\u6b21\u76f8\u52a0\u5373\u53ef\uff1b
\u4ee3\u7801\uff1a
public class Test {
public static void main(String[] args) {
int total=0;
for(int i=1;i<=100;i++){
total+=i;
}
System.out.println("1+2+3+...+100\u7684\u548c\u662f\uff1a"+total);
}
}
java\u8fd0\u884c\u8f93\u51fa\uff1a
1+2+3+...+100\u7684\u548c\u662f\uff1a5050
代码文本：
#include "stdio.h"
int main(int argc,char *argv[]){
int n,s;
for(s=0,n=1;n<101;s+=n*n++);//用for循环
printf("%16s: %d
","Using \'for\'",s);
s=0,n=1;//用while循环
while(n<101)
s+=n*n++;
printf("%16s: %d
","Using \'while\'",s);
s=0,n=1;//用do~while循环
do{
s+=n*n++;
}while(n<101);
printf("%16s: %d
","Using \'do~while\'",s);
return 0;
}
比较简单，供参考。
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