Non-negative Partial Sums
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1420 Accepted Submission(s): 544
Problem Description
You are given a sequence of n numbers a0,..., an-1. A cyclic shift by k positions (0<=k<=n-1) results in the following sequence: ak ak+1,..., an-1, a0, a1,..., ak-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,..., an-1 (-1000<=ai<=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
Source
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1 /* 2 题意： 3 刚刚又一道hdu的题目； 4 题意：10^6个数字，有10^6种形式。 5 如 a b c d , b c d a , c d a b, d a b c; 6 统计在所以情况中如果满足任意前i数和都>=0 的个数。 7 想了一下，思路有了。可以这样子。 8 任意前i数和都满足>=0，那么最小的是不是也就满足了呢? 9 肯定的。所以。题目可以转换为 10 以i为开头，长度为n的前提下，求最小值。 11 这样的话，只有q[head].sum - s[ i - n]; 12 13 */ 14 15 #include<iostream> 16 #include<stdio.h> 17 #include<cstring> 18 #include<cstdlib> 19 using namespace std; 20 21 int a[1000002]; 22 int s[2000002]; 23 typedef struct 24 { 25 int num; 26 int sum; 27 }Queue; 28 Queue tmp,q[2000004]; 29 30 int main() 31 { 32 int n,i,len,Num,tom; 33 int head,tail; 34 while(scanf("%d",&n)>0) 35 { 36 if(n==0)break; 37 for(i=1;i<=n;i++) 38 scanf("%d",&a[i]); 39 len=n*2; 40 for(i=n+1;i<=len;i++) 41 a[i]=a[i-n]; 42 for(s[0]=0,i=1;i<=len;i++) 43 s[i]=s[i-1]+a[i]; 44 head=0;tail= -1; Num=1; tom=0; 45 for(i=1;i<=len;i++) 46 { 47 tmp.num=++Num; 48 tmp.sum=s[i]; 49 while( head<=tail && q[tail].sum>tmp.sum ) tail --; 50 q[++tail]=tmp; 51 if( i>n ) 52 { 53 while( head<=tail && q[head].num+n<=i ) head++; 54 if( q[head].sum-s[i-n]>=0) tom++; 55 } 56 } 57 printf("%d ",tom); 58 } 59 return 0; 60 }