HDU 5776 sum(抽屉原理)
2016-08-01 09:00
316 查看
sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 851 Accepted Submission(s): 397
[align=left]Problem Description[/align]
Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO
[align=left]Input[/align]
The first line of the input has an integer T (1≤T≤10),
which represents the number of test cases.
For each test case, there are two lines:
1.The first line contains two positive integers n, m (1≤n≤100000,
1≤m≤5000).
2.The second line contains n positive integers x (1≤x≤100)
according to the sequence.
[align=left]Output[/align]
Output T lines, each line print a YES or NO.
[align=left]Sample Input[/align]
2
3 3
1 2 3
5 7
6 6 6 6 6
[align=left]Sample Output[/align]
YES
NO
[align=left]Source[/align]
BestCoder Round #85
[align=left]Recommend[/align]
wange2014 | We have carefully selected several similar problems for you: 5780 5779 5778 5777 5775
抽屉原理:如果现在有3个苹果,放进2个抽屉,那么至少有一个抽屉里面会有两个苹果
抽屉原理的运用
现在假设有一个正整数序列a1,a2,a3,a4.....an,试证明我们一定能够找到一段连续的序列和,让这个和是n的倍数,该命题的证明就用到了抽屉原理
我们可以先构造一个序列si=a1+a2+...ai
然后分别对于si取模,如果其中有一个sk%n==0,那么a1+a2+...+ak就一定是n的倍数(该种情况得证)
下面是上一种情况的反面,即任何一个sk对于n的余数都不为0
对于这种情况,我们可以如下考虑,因为si%n!=0
那么si%n的范围必然在1——(n-1),所以原序列si就产生了n个范围在1——(n-1)的余数,于是抽屉原理就来了,n个数放进n-1个盒子里面,必然至少有两个余数会重复,那么这两个sk1,sk2之差必然是n的倍数,
而sk1-sk2是一段连续的序列,那么原命题就得到了证明了
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int a[110000],cnt[110000]; int main() { int n,m,t; scanf("%d",&t); while(t--) { bool flag=false; memset(a,0,sizeof(a)); memset(cnt,0,sizeof(cnt)); scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); a[i]=(a[i]+a[i-1])%m; cnt[a[i]]++; if(cnt[a[i]]>1||!a[i]) flag=true; } if(flag==true) printf("YES\n"); else printf("NO\n"); } return 0; }
相关文章推荐
- hdu 5776 sum (抽屉原理)
- HDU 5776 sum(抽屉原理)
- HDU 5776 SUM (鸽巢原理 / 抽屉原理)
- hdu5776——sum
- hdu 5776 sum
- HDU - 5776 sum 前缀数组
- HDU 5776 sum (思维题)
- HDU 5776 sum(思维题+前缀和)
- HDU - 5776 sum/ 蓝桥杯省赛 K倍区间(抽屉原理)
- 思维题:抽屉原理 hdu 5776 sum & 51Nod 1103 N的倍数
- HDU 5776 sum (BestCoder Round #85 A) 简单前缀判断+水题
- HDU 5776 sum ( BC #85 1001)
- hdu-5776 sum(同余)
- HDU-5776 Sum
- HDU 5776 sum(水~)
- HDU 5776 sum (思维题)
- HDU 5776 sum(数学)
- HDU 5776 sum
- HDU 5776 sum
- HDU 5776 sum(前缀和+取模)