计算1到10之间奇数之和及偶数之和并输出
\u8ba1\u7b971\u523010\u4e4b\u95f4\u5947\u6570\u4e4b\u548c\u53ca\u5076\u6570\u4e4b\u548c\u5e76\u8f93\u51fa\u3002C\u8bed\u8a009.
int& z\u5427
10.
#include
int main(){
int a, b, c, i;
a=c=0;
for(i=0;i<=10;i+=2){/*\u8fd9\u91cc\u5e94\u8be5\u662f<=10\uff0c\u4e0d\u662f<10*/
a+=i;
b=i+1; /*\u586b\u7a7a*/
c+=b;
}
printf("\u5076\u6570\u4e4b\u548c\uff1d%d\n",a);
printf("\u5947\u6570\u4e4b\u548c\uff1d%d\n",c-11); /*\u6700\u540e\u4e00\u6b21\u5faa\u65f6,i=10;,b=11\uff0cc\u5b58\u653e\u5947\u6570\u548c\u4e14\u5728b=i+1\u540e\u9762\uff0c\u591a\u52a0\u4e8611*/
}
#include "stdio.h" void main() { int sum1=0,sum2=0,i; for(i=1;i<=10;i++) { if(i%2==0) sum2+=i;/*\u8ba1\u7b971\u523010\u95f4\u7684\u5076\u6570\u4e4b\u548c*/ else sum1+=i;/*\u8ba1\u7b971\u523010\u95f4\u7684\u5947\u6570\u4e4b\u548c*/ } printf("1\u523010\u95f4\u7684\u5947\u6570\u4e4b\u548c\uff1a%d\n",sum1); printf("1\u523010\u95f4\u7684\u5076\u6570\u4e4b\u548c\uff1a%d\n",sum2); }
温馨提示奇数之和=1+3+5+7+9=25
偶数之和=2+4+6+8+10=30
奇数;1+3+5+7+9=25
偶数；0+2+4+6+8+10=30
开始
输入X
m等于X除以2的余数
输出奇数，输出偶数
加和
具体框图参考一下教材
绛旓細鍋舵暟涔嬪拰=2+4+6+8+10=30
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