Little Hasan loves to play number games with his friends.One day they were playing a game where one of them will speak out a positive numberand the others have to tell the sum of its factors. The first one to say itcorrectly wins. After a while they got bored and wanted to try out a differentgame. Hassan then suggested about trying the reverse. That is, given a positivenumber S, they have to find a numberwhose factors add up to S. Realizingthat this task is tougher than the original task, Hasan came to you forhelp.  Luckily Hasan owns a portableprogrammable device and you have decided to burn a program to this device.Given the value of S as input to theprogram, it will output a number whose sum of factors equal to S.

Input

Each case of input will consist of a positive integer S<=1000. The last case is followedby a value of 0.

Output

Foreach case of input, there will be one line of output. It will be a positiveinteger whose sum of factors is equal to S.If there is more than one such integer, output the largest. If no such numberexists, output -1. Adhere to theformat shown in sample output.

Sample Input                            Output for Sample Input

ProblemSetter: Shamim Hafiz, Special Thanks: Sohel Hafiz

题意：输入一个正整数S，求一个最大的正整数N，使得N的所有因子的和是S

思路：统计1000内的数的因子和，求解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1005;

int s, sum[maxn];

void init() {
	memset(sum, 0, sizeof(sum));
	for (int i = 1; i < maxn; i++)
		for (int j = 1; j <= i; j++)
			if (i % j == 0)
				sum[i] += j;
}

int main() {
	int cas = 1;
	init();
	while (scanf("%d", &s) != EOF && s) {
		int flag = 0;
		printf("Case %d: ", cas++);
		for (int i = 1000; i >= 1; i--) {
			if (sum[i] == s) {
				printf("%d\n", i);
				flag = 1;
				break;
			}	
		}
		if (!flag)
			printf("-1\n");
	}	
	return 0;
}

UVA - 11728 Alternate Task （唯一分解定理）

原文：http://blog.csdn.net/u011345136/article/details/38817513