Project Euler - Problem 30

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#include #ios

本文介绍了一个C++程序，用于寻找特定范围内的五次幂和数，并计算这些数的总和。程序通过分解数字并计算各数字位上数值的五次幂之和来判断一个数是否为五次幂和数。

#include <iostream> #include <time.h> #include <math.h> #include <set> #include <list> using namespace std; int handle (int tmp) { int num = tmp; int a[6] = {0}; if( tmp/100000 > 0 ) { a[0] = tmp/100000; tmp = tmp - a[0] * 100000; } else a[0] = 0; if ( tmp/10000 > 0 ) { a[1] = tmp/10000; tmp = tmp - a[1] * 10000; } else a[1] = 0; if ( tmp/1000 > 0 ) { a[2] = tmp/1000; tmp = tmp - a[2] * 1000; } else a[2] = 0; a[3] = tmp/100; tmp = tmp - a[3] * 100; a[4] = tmp/10; tmp = tmp - a[4] * 10; a[5] = tmp; int sum = 0; int index=0; if(a[0]==0) { index++; if(a[1]==0) index++; } for(int i=index; i!=6; i++) { sum += a[i]*a[i]*a[i]*a[i]*a[i]; //pow((double)a[i],5); 比用pow函数快了1秒 } if(num == sum) return num; return 0; } int main() { double start = clock(),end(0); int sum = 0; for(int n = 1000; n<300000; n++) { int result = handle(n); if( 0!=result) { sum += result; cout <<"number : " << result<<endl; } } cout <<"the sum : " << sum; end = clock(); double result = (end - start)/1000; cout.setf(ios::fixed); cout <<endl<< "3.00GHz's run time = " << result << "s."<<endl; return 0; }