Non-negative Partial Sums
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Problem Description You are given a sequence of n numbers a
0,..., 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, a
0, a
1,..., 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<=10
6), the number of integers in the sequence. The second contains n integers a
0,..., 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
#include
#include
using namespace std; const int maxn = 1000010; int a[maxn*2]; int q[maxn*2]; int main() { int n; while(scanf("%d", &n) && n) { for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); a[i+n] = a[i]; } int cnt = 0; for(int i = 1; i <= 2*n; i++) a[i] += a[i-1]; int front = 0, rear = -1; for(int i = 1; i < 2*n; i++) { while(front <= rear && i - q[front] >= n) front++; while(front <= rear && a[i] <= a[q[rear]]) rear--; q[++rear] = i; if(i >= n && a[q[front]] - a[i-n] >= 0) cnt++; } printf("%d\n", cnt); } return 0; }