Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n ≤ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
The first line contains two integers k and n (1 ≤ n ≤ k ≤ 2 000) — the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 ≤ ai ≤ 2 000) — jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 ≤ bj ≤ 4 000 000) — the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
4 1-5 5 0 2010
3
2 2-2000 -20003998000 4000000
1
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.
先前缀和排一遍序，那么n个某时刻后的总分数也会按照所排顺序出现。
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))
#define pos first
#define index second
#define mp make_pair
using namespace std;
const int inf=1e9+10;
const ll llinf=1e16+10;
const int maxn=2e3+10;
const int maxm=1e2+10;
const int mod=1e9+7;
int k,n;
int score[maxn];
int sum[maxn];
int reb[maxn];
int main()
{
//freopen("in.txt","r",stdin);
while(~scanf("%d %d",&k,&n))
{
set<int> S;
sum[0]=0;
for(int i=1;i<=k;i++)
{
scanf("%d",&score[i]);
sum[i]=sum[i-1]+score[i];
S.insert(sum[i]);
}
sort(sum+1,sum+k+1);
for(int i=1;i<=n;i++)
scanf("%d",&reb[i]);
sort(reb+1,reb+1+n);
int ans=0;
for(int i=1;i<=k;i++)
{
if(i==1||sum[i]!=sum[i-1])
{
int t=reb[1]-sum[i];
int flag=1;
for(int j=2;j<=n;j++)
{
if(S.find(reb[j]-t)==S.end())
{flag=0;break;}
}
ans+=flag;
}
}
printf("%d\n",ans);
}
return 0;
}
来源：https://www.cnblogs.com/mgz-/p/7168345.html