2016 Multi-University Training Contest 5 Divide the Sequence
Divide the Sequence
题意:
给你一个序列A,问你最多能够分成多少个连续子序列,使得每个子序列的所有前缀和均不小于0
题解:
这题是比赛时候的水题,但我比的时候也就做出这一题, = =
首先我想的是把他们前缀和求出来,之后试了下样例,一点鸟用都没有,那正着不行就倒着试下呗,之后发现这样有用,我先想到的思路是求后缀和,只要>=0就ans++,但那时队友举出了反例,比如2 1 -3 3的时候应该是2,我的算法就是4,那接着马上就能想到如果后缀和大于0,那么就要把他赋为0,之后举了几组例子,没问题,写了几分钟,之后1A。
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const ll LINF=0x3f3f3f3f3f3f3f3f;
#define PI(A) cout<<A<<endl
#define SI(N) cin>>N
#define SII(N,M) cin>>N>>M
#define cle(a,val) memset(a,(val),sizeof(a))
#define rep(i,b) for(int i=0;i<(b);i++)
#define Rep(i,a,b) for(int i=(a);i<=(b);i++)
#define reRep(i,a,b) for(int i=(a);i>=(b);i--)
#define dbg(x) cout <<#x<<" = "<<x<<endl
#define PIar(a,n) rep(i,n)cout<<a[i]<<" ";cout<<endl;
#define PIarr(a,n,m) rep(aa,n){rep(bb, m)cout<<a[aa][bb]<<" ";cout<<endl;}
const double EPS= 1e-9 ;
/* ///////////////////////// C o d i n g S p a c e ///////////////////////// */
const int MAXN= 1000000 + 9 ;
ll a[MAXN];
int N;
int main()
{
iostream::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
while(SI(N))
{
ll ans=0;
rep(i,N) SI(a[i]);
ll sum=0;
reRep(i,N-1,0)
{
sum+=a[i];
if (sum>=0) sum=0;
if (sum==0) ans++;
}
PI(ans);
}
return 0;
}
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