= 0) cheng_ji = cheng_ji*i; else sum_ou +=i; } printf("%d %d", sum_ou, cheng_ji); return 0;}include "stdio.h"void main(){int sum1=0,sum2=1;for (int i=1;i<=10;i++){sum1+=2*i;sum2*=(2*i-1);}printf("%d,%d",sum1,sum2);retu...
include <stdio.h> int main(){ int i,j,sum=0;for(i=1,j=10;i<10;i++,j++){ sum=sum+i+j;printf("i=%d,j=%d,sum=%d\n",i,j,sum);} } 结果：i=1,j=10,sum=11 i=2,j=11,sum=24 i=3,j=12,sum=39 i=4,j=13,sum=56 i=5,j=14,sum=75 i=6,j=15,sum=...
for(i=1;i<=100;i+=2)sum+=i;printf("奇数和：%d\n",sum);sum=0;for(i=2;i<=100;i+=2;sum+=i;printf("偶数和：%d\n",sum);C语言中编程计算1至100以内的奇数和偶数并分别求和 include<stdio.h> void main(){ int i,sumA=0,sumB=0;for(i=1;i<=100;i++){ if(i%2!
include<stdio.h> void main(){ int i,sum=0;for(i=1;i<=10;i++)sum+=i*i;printf("sum=%d\n",sum);i=1;sum=0;while(i<11){ sum+=i*i;i++;};printf("sum=%d\n",sum);i=1;sum=0;do { sum+=i*i;i++;}while (i<11);printf("sum=%d\n",sum);} for...
include <stdio.h>int main(void) {int i,sum;for(i=1,sum=0;i<101;++i){sum+=i;}printf("%d\n",sum);return 0;} include
include<stdio.h> void main(){ int i,j,sum,f,k,n;sum=0;f=1;printf("请输入k次幂和n个数\n");scanf("%d%d",&k,&n);for ( i=1;i<=n;i++){ for (j=1;j<=k;j++){ f=f*i;} printf("%3d",f);sum=sum+f;f=1;} printf("\n表达式结果为：\n");printf("%d\...
这个程序的功能是算1+2+3+4+5；for循环的过程 开始 i=1 因为i小于6 执行sum+i=0+1=1，再把1赋给sum，此时sum=1, i++=i+1=1+1=2,；i=2 因为i小于6 执行sum+i=1+2=3，再把3赋给sum，此时sum=3, i++=i+1=2+1=3,；依次递推 i=5 因为i小于6 执行sum+i=10...
include <stdio.h> int main(void){ int i;int j = 0;int am = 0;for (i = 1; i <= 100; i += 2){ j = j + i;if (i% 2 == 1) //这里是i%2，实际上这个条件没必要，因为i+=2了，肯定是奇数 ++am;} printf("请输出1-100 中所有奇数的和= %d\n", j);printf...
include "stdio.h"void main(){long sum=0,temp=1;int i,j;for(i=1;i<=10;i++){for(j=1;j<=i;j++){temp=temp*j;}sum=sum+temp;temp=1;}printf("%ld",sum);}include "stdio.h"void main(void){int i,j,k,sum;for(sum=0,i=1;i<11;i++){for(k=1,j=i;j>1;k*...
}}cout<<"100以内的基数和为:"<<s_j<<"，偶数积为"<<s_o<<endl;}include"stdio.h"include"math.h"voidmain(){inti,sum=0;longdoublek=1;for(i=1;i<=100;i++){if(i%2==0)/*判断偶数*/k*=i;elsesum+=i;}printf("0-100间奇数之和为%d，偶数之积为%ld"，sum，k)；}in...