#include<stdio.h> #include<math.h> int main() { int i,x,y,last=1; //变量last保存求X的Y次方过程中的部分积的后三位 printf("Input X and Y(X**Y):"); scanf("%d**%d",&x,&y); for(i=1;i<=y;i++) //X自乘Y次 last=last*x00; //将last乘X后对1000取模，即求积的后三位 printf("The last 3 dights of %d**%d is :%d/n",x,y,last00);//%运算符的意义为取余 return 0; //用00运算式取数值的后三位 } /* output: ----------------------------------------------------------------------- Input X and Y(X**Y):13**20 The last 3 dights of 13**20 is :801 Process returned 0 (0x0) execution time : 4.641 s Press any key to continue. ----------------------------------------------------------------------- */
