题意：
给定一个由n个整数组成的整数序列，可以滚动，滚动的意思就是前面k个数放到序列末尾去。问有几种滚动方法使得前面任意个数的和>=0.
思路：
先根据原来的数列求sum数组，找到最低点，然后再以最低点为始点，求解题目答案，（每求解一始点i，符合要求的条件为：sum[i]>=minx,[minx是i<x<=n中的最小值],之所以不用考虑前面的，就是因为我们的预处理是的所有的x<i的sum[x]的值满足sum[x]>=sum[i]）
代码如下：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#include <set>
#include <map> #define M 1000005
#define mod 1000000007
#define INF 0x7fffffff
#define eps 1e-8
#define LL long long
#define LLU unsigned long long using namespace std; int a[M], n, mini;
LL sum[M], minx;
int main ()
{
while(scanf("%d", &n) && n)
{
minx = INF;
sum[0] = 0;
for(int i= 1; i <= n; ++i)
{
scanf("%d", &a[i]);
sum[i] = sum[i-1]+a[i];
if(minx>sum[i])
{
minx = sum[i];
mini = i;
}
}
sum[0] = 0;
int c = 0;
for(int i = mini+1; i <= n; ++i)
sum[++c] = sum[c-1]+a[i];
for(int i = 1; i <= mini; ++i)
sum[++c] = sum[c-1]+a[i];
minx = sum[n];
int ans = 0;
for(int i = n-1; i >= 0; --i)
{
if(sum[i]<=minx)
{
ans += 1;
minx = sum[i];
}
}
printf("%d\n", ans);
}
return 0;
}
单调队列的思路是比较简单的，
就是确定一个始点以后，然后在队列中找最小的值，满足要求的条件为：minx-sum[i]>=0
代码如下：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#include <set>
#include <map> #define M 1000005
#define mod 1000000007
#define INF 0x7fffffff
#define eps 1e-8
#define LL long long
#define LLU unsigned long long using namespace std;
int n, head, rear, a[M];
LL sum[2*M], deq[2*M];
void insert(int x)
{
while(head<=rear && sum[deq[rear]]>=sum[x]) --rear;
deq[++rear] = x;
}
LL push(int x)
{
while(deq[head]<=x-n) ++head;
return sum[deq[head]];
}
int main ()
{
while(scanf("%d", &n) && n)
{
head = 1; rear = 0;
for(int i = 1; i <= n; ++i)
scanf("%d", &a[i]);
sum[0] = 0;
for(int i = 1; i <= n; ++i)
sum[i] = sum[i-1]+a[i];
for(int i = n+1; i <= 2*n; ++i)
sum[i] = sum[i-1]+a[i-n];
for(int i = 1; i < n; ++i)
insert(i);
int ans = 0;
for(int i = n; i < 2*n; ++i)
{
insert(i);
if(push(i)-sum[i-n]>=0) ++ans;
}
printf("%d\n", ans);
}
return 0;
}