c语言编程计算(a*b)%c c语言编程：计算 a / b
c\u8bed\u8a00\u7f16\u7a0b\u8ba1\u7b97(a*b)%cint a,b,c;scanf("%d%d%d",&a,&b,&c);printf("%d",a*b%c);
#include int main(){ int a,b,c; do{ scanf("%d%d", &a,&b); fflush(stdin); }while(a10000||b>10000); if(b==0)printf("Division by zero.\n"); else { c = a/b; printf("%d\n", c); } return 0;}
直接二分幂取模，so easy！#include<stdio.h>
int pow_mod(int a,int b,int c)
{
int t;
if(b==0)
return 1;
if(b==1)
return a%c;
t=pow(a,b/2,c);
t=t*t%c;
if(b&1==1)
t=t*a%c;
return t;
}
int main()
{
int a,b,c;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
printf("%d\n",pow_mod(a,b,c));
}
return 0;
}
应该就是这样，现打的，没调试，应该过的了，要不行就把int改成__int64
大数问题，只能改变算法 ，主要思想就是将数字转换为字符，字符串处理就简单了。
#include<stdio.h>
main()
{
int a,b,c,s,h;
printf("输入abc：") //输入数据
scanf("%d%d%d",&a&b&c)
s=a*b;//计算
h=s%c;
printf("(a*b)%c=%d",h);//输出结果
}
这个应该属于大数的范畴，可以考虑改下算法 ；主要思想就是讲数字转换为字符来处理
#include<stdio.h>
#include<stdlib.h>
void main()
{
__int64 a,b,c;
while(scanf("%I64d %I64d %I64d",&a,&b,&c)!=EOF)
{
printf("%I64d %I64d %I64d\n",a,b,c);
printf("%I64d\n",((a%c)*(b%c))%c);
}
}
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绛旓細灏辨暟瀛︿笂鏉ヨ涓鏍 1/a/b/c = ((1/a)/b)/c = 1/(a*b*c)浣嗘槸鍦╟閲岄潰绮剧‘搴︿笂浼氬嚭鐜颁笉鍚岀粨鏋 姣斿閮芥槸int 100/200/0.001 = (100/200)/0.0001 = 0/0.0001 = 0 浣嗘槸100/(200*0.001) = 100/0.2 = 500
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绛旓細include <stdio.h>int main(void){ void output(int num[][3]); int i, j; int a[3][3], b[3][3], c[3][3]; puts("璇疯緭鍏ョ煩闃礱 : "); for (i = 0; i < 3; ++i) { for (j = 0; j < 3; ++j) scanf("%d", a[i] + j); } puts("璇疯緭鍏ョ煩闃礲 : ...
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