hdu 5890 Eighty seven

最新推荐文章于 2019-05-16 21:27:33 发布

原创最新推荐文章于 2019-05-16 21:27:33 发布 · 624 阅读

0 ·

CC 4.0 BY-SA版权

dp 专栏收录该内容

31 篇文章

订阅专栏

本文介绍了一个有趣的数学挑战问题，通过给定的一组数字，在每轮移除不多于3个数字的情况下，判断是否能从剩余数字中选出10个数，使得这些数的总和等于87。文章提供了详细的解题思路和使用bitset优化后的DP算法实现。

Eighty seven

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 243 Accepted Submission(s): 74

Problem Description

Mr. Fib is a mathematics teacher of a primary school. In the next lesson, he is planning to teach children how to add numbers up. Before the class, he will prepare

N cards with numbers. The number on the

i -th card is

ai . In class, each turn he will remove no more than

3 cards and let students choose any ten cards, the sum of the numbers on which is

87 . After each turn the removed cards will be put back to their position. Now, he wants to know if there is at least one solution of each turn. Can you help him?

Input

The first line of input contains an integer

t (t≤5) , the number of test cases.

t test cases follow.
For each test case, the first line consists an integer

N(N≤50) .
The second line contains

N non-negative integers

a1,a2,...,aN . The

i -th number represents the number on the

i -th card. The third line consists an integer

Q(Q≤100000) . Each line of the next

Q lines contains three integers

i,j,k , representing Mr.Fib will remove the

i -th,

j -th, and

k -th cards in this turn. A question may degenerate while

i=j ,

i=k or

j=k .

Output

For each turn of each case, output 'Yes' if there exists at least one solution, otherwise output 'No'.

Sample Input

  
   1
12
1 2 3 4 5 6 7 8 9 42 21 22
10
1 2 3
3 4 5
2 3 2
10 10 10
10 11 11
10 1 1
1 2 10
1 11 12
1 10 10
11 11 12

Sample Output

  
   No
No
No
Yes
No
Yes
No
No
Yes
Yes

Source

2016 ACM/ICPC Asia Regional Qingdao Online

Recommend

wange2014 | We have carefully selected several similar problems for you: 5891 5889 5887 5886 5885

题意：给你n个数，m次询问，每次询问给你三个数，问在给出的数里面存不存在任意取10个（不包含这三个数）的数的和==87。

思路：用bitset暴力dp，时间卡着过，暴力打表用dp[i][j][k]表示存不存在，然后就是用bitset优化0 1背包。剩下的看代码：

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<functional>
typedef long long LL;
using namespace std;
#define inf 0x3f3f3f3f
#define maxn 55
typedef long long LL;
int a[maxn], dp[maxn][maxn][maxn];
bitset<90>s[11];
int n;
bool solve(int x, int y, int z){
	for (int i = 0; i<11; i++)
		s[i].reset();
	s[0][0] = 1;
	for (int i = 1; i <= n; i++){
		if (i == x || i == y || i == z || a[i]>87)
			continue;
		for (int j = 10; j>0; j--){
			s[j] |= s[j - 1] << a[i];
		}
	}
	if (s[10][87])
		return 1;
	else
		return 0;
}
int main(){
	int t;
	scanf("%d", &t);
	while (t--){
		scanf("%d", &n);
		for (int i = 1; i <= n; i++){
			scanf("%d", &a[i]);
		}
		for (int i = 1; i <= n; i++){
			for (int j = i; j <= n; j++){
				for (int k = j; k <= n; k++){
					dp[i][j][k] = solve(i, j, k);
				}
			}
		}
		int q;
		scanf("%d", &q);
		while (q--){
			int b[3];
			scanf("%d%d%d", &b[0], &b[1], &b[2]);
			sort(b, b + 3);
			if (dp[b[0]][b[1]][b[2]])
				printf("Yes\n");
			else
				printf("No\n");
		}
	}
}