UVa 11728 Alternate Task (枚举)

题意：给定一个 n，求一个最大正整数 N 使得 N 的所有正因数和等于 n。

析：对于任何数一个 n，它的所有正因子都是大于等于本身的，因为 n 本身就是自己的正因数，这样的就可以直接暴力了，答案肯定是在 1 ~ n 范围内。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define aLL 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int maxn = 1000 + 20;
const int maxm = 76543;
const int mod = 1e9 + 9;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }

int ans[maxn];

int main(){
  ms(ans, -1);
  for(int i = 1; i <= 1000; ++i){
    int t = sqrt(i + 1.);
    int sum = 0;
    for(int j = 1; j < t; ++j)  if(i % j == 0)
      sum += j + i / j;
    if(i % t == 0){
      sum += t;
      if(t != i / t)  sum += i / t;
    }
    if(sum < maxn)  ans[sum] = i;
  }
  int kase = 0;
  while(scanf("%d", &n) == 1 && n)
    printf("Case %d: %d\n", ++kase, ans[n]);
  return 0;
}