文章目錄

• • 題目描述
  • 題意：
  • 題解：

傳送

時(shí)間限制：C/C++ 2秒，其他語言4秒 空間限制：C/C++ 131072K，其他語言262144K
64bit IO Format: %lld

題目描述

Listening to the music is relax, but for obsessive(強(qiáng)迫癥), 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.

輸入描述:

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.

輸出描述:

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
輸入

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

輸出

NO YES YES

說明
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

題意：

給你n個(gè)數(shù)，這些數(shù)自由組合能不能湊出3600的倍數(shù)

題解：

我一開始想到的是前綴和，后來感覺dp最直接
dp[x]=1表示能組成x這個(gè)數(shù)
dp = 0表示組不了
cnt是中間數(shù)組，暫時(shí)存儲(chǔ)本輪的數(shù)值
因?yàn)榍竽懿荒芙M成3600，可以用mod，3600的倍數(shù)mod后都是0，直接求dp[0]是否等于1
每讀取一個(gè)a，就把a(bǔ)與之前所求的值進(jìn)行相加存在cnt里，然后再給dp[]，cnt就是工具人

#include<bits/stdc++.h> #define mem(a) memset(a,0,sizeof(a))using namespace std; const int maxn=1e5+3; bool dp[maxn],cnt[maxn]; const int mod=3600; int main() {int t,n;scanf("%d",&t);while(t--){mem(dp);mem(cnt);scanf("%d",&n);for(int i=1;i<=n;i++){int a;cin>>a;a%=3600;if(!dp[0]){for(int j=0;j<3600;j++){if(dp[j]>0||j==0){cnt[(a+j)%3600]=1;}}for(int j=0;j<3600;j++){if(cnt[j])dp[j]=1;if(cnt[j]==1)cnt[j]=0;}// mem(cnt);}}if(!dp[0])cout<<"NO"<<endl;else cout<<"YES"<<endl;}return 0; }

有個(gè)很玄學(xué)的地方我把讀入n放在兩個(gè)mem之前，數(shù)據(jù)就過了一半，放后面就ac了，不知道為什么
看來卡時(shí)間卡的太緊了（笑哭）

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