求x的y的幂次方的最后3位数
求x的y的幂次方的最后3位数
程序代码如下:
/* 2017年3月12日14:07:05 功能:程序求x的y的幂次方的最后3位数 */ #include"stdio.h" void fun(int); int main() { int x, y; int x_power_y = 1; printf("please int two number x and y :"); scanf("%d %d", &x, &y); for (int i = 0; i < y; i++) { x_power_y *= x; } if (x_power_y < 999) { printf("the last three number of the x power y is %d ", x_power_y); } else { fun(x_power_y); //大于999,进入调用函数 } } void fun(int x_power_y) { int first = x_power_y % 10; //此时求得是最后一位上的数值 int second = (x_power_y % 100 - first * 1) / 10; //x_power_y % 100是最后两位上的数值 int thrid = (x_power_y % 1000 - second * 10 - first * 1) / 100; printf("the last three number of the x power y is %d ", thrid*100+second*10+first); } /* 总结; 在VC++6.0中显示的结果: —————————————————— please int two number x and y :11 3 the last three number of the x power y is 331 —————————————————— */