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UVALive 6896 Eureka Theorem(完全背包)

2015-12-15 01:20 344 查看
题意:

问是否恰好3个三角形数的和为n≤1000,三角形数为x=i∗(i+1)2

分析:

数可以重复,完全背包搞搞就好

代码:

//
//  Created by TaoSama on 2015-12-14
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n;
bool dp[1005][4];

vector<int> num;
void gao() {
int sum = 0;
num.push_back(0);
for(int i = 1; i <= 1000; ++i) {
sum += i;
if(sum > 1000) break;
num.push_back(sum);
}
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
gao();
while(t--) {
scanf("%d", &n);
memset(dp, false, sizeof dp);
dp[0][0] = true;
for(int i = 1; i < num.size(); ++i)
for(int j = num[i]; j <= n; ++j)
for(int k = 1; k <= 3; ++k)
dp[j][k] |= dp[j - num[i]][k - 1];
printf("%d\n", dp
[3]);
}
return 0;
}
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标签:  dp