hdu 4193 Non-negative Partial Sums 单调队列

Problem Description

You are given a sequence of n numbers a₀,..., a_n-1. A cyclic shift by k positions (0<=k<=n-1) results in the following sequence: a_k a_k+1,..., a_n-1, a₀, a₁,..., a_k-1. How many of the n cyclic shifts satisfy the condition that the sum of the first i numbers is greater than or equal to zero for all i with 1<=i<=n?

Input

Each test case consists of two lines. The first contains the number n (1<=n<=10⁶), the number of integers in the sequence. The second contains n integers a₀,..., a_n-1 (-1000<=a_i<=1000) representing the sequence of numbers. The input will finish with a line containing 0.

Output

For each test case, print one line with the number of cyclic shifts of the given sequence which satisfy the condition stated above.

Sample Input

3 2 2 1 3 -1 1 1 1 -1 0

Sample Output

3 2 0

方法 一

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
using namespace std;

int sum[2000006];
int sk[2000006];

int main( )
{
    int i,j,top,hid,N,val,res;
    while( scanf("%d",&N) && N )
    {
        for( i = 1; i <= N; i++ )
        {
            scanf("%d",&sum[i]);
            sum[i+N] = sum[i];
        }
        for( i = 2; i <= N*2; i++ )
            sum[i] += sum[i-1];
        top = hid = 1; sk[1] = 1,res = 0;
        for( i = 2; i <= N*2; i++ )
        {
            while( top <= hid && sum[i] < sum[sk[hid]] )  hid--;
            while( top <= hid && i > N && sk[top] + N < i ) top++; 
                                                   sk[++hid] = i;
            if(  i > N && sum[sk[top]] - sum[i-N] >= 0 ) res++;
        }
            printf("%d\n",res);
    }
    //system("pause");
    return 0;
}


方法 二

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

struct point
{
      int w,num;
}q[1000006];
int  val[1000006];
int  Min[1000006];

bool cmp( point a,point b )
{
      return a.w < b.w;
}

int min( int a,int b )
{
    return a>b?b:a;
}

int main(  )
{
      int N,i,k,ans;
      while( scanf("%d",&N) && N )
      {
            val[0] = 0;
            Min[0] = 1000000001;
            for( i = 1; i <= N; i++ )
            {
                scanf("%d",&val[i]);
                val[i] += val[i-1];
                Min[i] = min( Min[i-1],val[i] );
                q[i].w = val[i];  q[i].num = i;
            }
            sort( &q[1],&q[1] + N,cmp );
            k = 1;   ans = 0;
            for( i  = 1; i <= N; i++ )
            if( val[i] < 0 )
                break;
            if( i == N + 1  ) ans++;
            for( i = 1; i < N; i++ )
            {
                int sum = val[N] - val[i];
                if( sum < 0 ) continue;
                int sta = i + 1;
                while( q[k].num < sta )       k++;
                if( k >= N + 1 ) break;
                if( q[k].w - val[i] < 0 ) continue;
                if( Min[i] + sum >= 0 ) ans++;
            }
            printf("%d\n",ans);
        }
    //system("pause");
    return 0;
}