问题:设计一个程序,求出712的729次方的最后4位数是多少?
大整数相乘问题,不能简单的乘,早就溢出了.所以要另辟蹊径.在相乘时每次取出后四位其余丢弃,不影响最后的四位数.
源代码如下:
#include <iostream> #include <iomanip> using namespace std; int main() { int value = 1; for(int i=1;i<=729;++i) { value*=712; value%=10000; } //设置输出格式,4位,用0填充. cout<<setw(4)<<setfill('0')<<value<<endl; system("pause"); return 0; }
如果我们注意到729 = 3^6,可以减少循环次数.
#include<stdio.h> int main() { unsigned long long a3 = (712*712*712)%10000; unsigned long long a9 = (a3 *a3*a3)%10000; unsigned long long a27 = (a9 *a9*a9)%10000; unsigned long long a81 = (a27 *a27*a27)%10000; unsigned long long a243 = (a81 *a81*a81)%10000; unsigned long long a729 = (a243 *a243*a243)%10000; printf ("result=%d \n",a729); getchar(); return 0; }
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