<span>牛客算法周周练2 - B -Music Problem （背包+bitset优化）</span>

牛客算法周周练2

链接：https://ac.nowcoder.com/acm/contest/5203/B
来源：牛客网

Music Problem

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld

题目描述

Listening to the music is relax, but for obsessive(强迫症), it may be unbearable.

HH is an obsessive, he only start to listen to music at 12:00:00, and he will never stop unless the song he is listening ends at integral points (both minute and second are 0 ), that is, he can stop listen at 13:00:00 or 14:00:00,but he can't stop at 13:01:03 or 13:01:00, since 13:01:03 and 13:01：00 are not an integer hour time.

Now give you the length of some songs, tell HH whether it's possible to choose some songs so he can stop listen at an integral point, or tell him it's impossible.

Every song can be chosen at most once.

输入描述:

 The first line contains an positive integer T(1≤T≤60), represents there are T test cases. 
 For each test case: 
 The first line contains an integer n(1≤n≤105), indicating there are n songs. 
 The second line contains n integers a1,a2…an (1≤ai≤109 ), the ith integer ai indicates the ith song lasts ai seconds.

输出描述:

For each test case, output one line "YES" (without quotes) if HH is possible to stop listen at an integral point, and "NO" (without quotes) otherwise.

示例1

输入

[复制](javascript:void(0)😉

3
3
2000 1000 3000
3
2000 3000 1600
2
5400 1800

输出

[复制](javascript:void(0)😉

NO
YES
YES

说明

In the first example it's impossible to stop at an integral point.
In the second example if we choose the first and the third songs, they cost 3600 seconds in total, so HH can stop at 13:00:00
In the third example if we choose the first and the second songs, they cost 7200 seconds in total, so HH can stop at 14:00:00

题意：

现在的时间是\(12:00:00\)，你有\(\mathit n\)个歌曲，第\(\mathit i\)个歌曲时长\(a_i\)秒，问是否能选择一些歌曲且只听一遍，在结束时也是整小时。

\(1\le1 a_i\leq 1e5\)

思路：

显然可以让\(a_i\)先对\(3600\)取模，然后用一个\(bitset<7201> dp\)维护\(dp\)信息

\(dp_i\)代表可以在第\(\mathit i\)秒停止。

然后对于每一个\(a_i\)，

用bitset的位运算来更新dp数组：

a |= a << x;
a |= a >> 3600;

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <unordered_map>
// #include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x)  if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 1
int n;
// int a[maxn];
bitset<8000> a;
bitset<8000> b;
int main()
{
#if DEBUG_Switch
    freopen("C:\\code\\input.txt", "r", stdin);
#endif
    //freopen("C:\\code\\output.txt","r",stdin);
    int t;
    t = readint();
    int x;
    while (t--)
    {
        a.reset();
        n = readint();
        a[0] = 1;

        repd(i, 1, n)
        {
            x = readint();
            x %= 3600;
            a |= a << x;
            a |= a >> 3600;
        }
       if (a[3600])
        {
            printf("YES\n");
        } else
        {
            printf("NO\n");
        }


    }

    return 0;
}