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
1 102 1000 0
|
Case 1: 1 Case 2: 101 Case 3: -1
|
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; }