用三种循环结构实现求1+2-3+4-5+~+101,java提供的两种循环控制语句是什么? 利用循环结构实现求1+2+3+…..+100的和,并将运算的...

\u7528\u4e09\u79cd\u5faa\u73af\u7ed3\u6784\u6c42\uff1a1+2+3++---+100\u53ca1*2*---*100\u7684\u503c

#include
int main()
{
int i,sum=0;
double mul=1;
for(i=1;i<=100;i++)
{
sum+=i;
mul*=i;
//\u4e00\u79cd
}
printf("%d %lf\n",sum,mul);
i=1;
sum=0;
mul=1;
while(i<=100)
{
sum+=i;
mul*=i;
i++;
//\u4e24\u79cd
}
i=1;
printf("%d %lf\n",sum,mul);
sum=0;
mul=1;
do
{
sum+=i;
mul*=i;
i++;
//\u4e09\u79cd
}while(i<=100);
printf("%d %lf\n",sum,mul);
return 0;
}
\u90a3\u4e2a1*2*3*...100\u8fd9\u4e2a\u503c\u4e5f\u591f\u5927\u7684

#include stadio.h;
void main(){
int i,sum=0;
for(i=1;i<=100;i++){
sum=sum+i;}
printf("1+2+3+.......+100=",&sum);
}

\u6269\u5c55\u8d44\u6599
C\u8bed\u8a00\u7684\u8fd0\u7b97\u7b26\u4e3b\u8981\u7528\u4e8e\u6784\u6210\u8868\u8fbe\u5f0f\uff0c\u540c\u4e00\u4e2a\u7b26\u53f7\u5728\u4e0d\u540c\u7684\u8868\u8fbe\u5f0f\u4e2d\uff0c\u5176\u4f5c\u7528\u5e76\u4e0d\u4e00\u81f4\u3002\u4e0b\u9762\u6309\u8ba1\u7b97\u7684\u4f18\u5148\u987a\u5e8f\uff0c\u5206\u522b\u8bf4\u660e\u4e0d\u540c\u4f5c\u7528\u7684\u8868\u8fbe\u5f0f\u3002\u9700\u8981\u7279\u522b\u6307\u51fa\uff0c\u5728C\u8bed\u8a00\u6807\u51c6\u4e2d\uff0c\u5e76\u6ca1\u6709\u7ed3\u5408\u6027\u7684\u8bf4\u6cd5\u3002
\u76f8\u540c\u4f18\u5148\u7ea7\u8fd0\u7b97\u7b26\uff0c\u4ece\u5de6\u81f3\u53f3\u4f9d\u6b21\u8fd0\u7b97\u3002\u6ce8\u610f\u540e\u7f00\u8fd0\u7b97\u4f18\u5148\u7ea7\u9ad8\u4e8e\u524d\u7f00\u3002\u56e0\u6b64++i++\u5e94\u89e3\u91ca\u4e3a++(i++)\u3002
\u800c\u4e0e\u6216\u975e\u7684\u8fd0\u7b97\u4f18\u5148\u7ea7\u90fd\u4e0d\u4e00\u6837\uff0c\u56e0\u6b64a && b || b && c\u89e3\u91ca\u4e3a(a && b) || (b && c)
\u5408\u7406\u4f7f\u7528\u4f18\u5148\u7ea7\u53ef\u4ee5\u6781\u5927\u7b80\u5316\u8868\u8fbe\u5f0f\u3002

int i1=0;
for(int i=1;i<102;i++){
if (i%2==1){
i1+=i;
}else{
i1-=i;
}

}sysout(i1)
//第一种,for循环
i1=0;
while(i<102){
if (i%2==1){
i1+=i;
}else{
i1-=i;
}
}
//第二种,while循环
sysout(i1)
int i1=0;
do{
if (i%2==1){
i1+=i;
}else{
i1-=i;
}
}while(i<102);
//第三种,do...while()循环

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