?題干：

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?NN?cards with numbers. The number on the?ii-th card is?aiai. In class, each turn he will remove no more than?33?cards and let students choose any ten cards, the sum of the numbers on which is?8787. 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)t?(t≤5), the number of test cases.?tttest cases follow.?
For each test case, the first line consists an integer?N(N≤50)N(N≤50).?
The second line contains?NN?non-negative integers?a1,a2,...,aNa1,a2,...,aN. The?ii-th number represents the number on the?ii-th card. The third line consists an integer?Q(Q≤100000)Q(Q≤100000). Each line of the next?QQ?lines contains three integers?i,j,ki,j,k, representing Mr.Fib will remove the?ii-th,?jj-th, and?kk-th cards in this turn. A question may degenerate while?i=ji=j,?i=ki=k?or?j=kj=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

題目大意：

50個(gè)數(shù)，10W個(gè)詢(xún)問(wèn)，每次問(wèn)刪掉第i,j,k個(gè)數(shù)后，是否存在一種選10個(gè)數(shù)和為87的方案，只需要輸出 ’Yes’ 或者 ’No’

解題報(bào)告：

? ?直接可行性背包，然后bitset優(yōu)化一下就可以了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #include<bitset> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; bitset<111> bs[111]; int a[MAX],n; bool dp[111][111][111]; bool ok(int x,int y,int z) {for(int i = 1; i<=n; i++) bs[i].reset();bs[0][0]=1;for(int i = 1; i<=n; i++) {if(i == x || i == y || i == z) continue;for(int j = 10; j>=1; j--) {bs[j] |= (bs[j-1] << a[i]);}}return bs[10][87]; } int main() {int t;cin>>t;while(t--) {scanf("%d",&n);memset(dp,0,sizeof dp);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]=0;if(ok(i,j,k)) dp[i][j][k] = 1;}}}int q,x[4];scanf("%d",&q);while(q--) {for(int i = 1; i<=3; i++) scanf("%d",x+i);sort(x+1,x+4);if(dp[x[1]][x[2]][x[3]] == 1) puts("Yes");else puts("No");}}return 0 ; }

?

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