CodeForces - 659G Fence Divercity （DP）好题

CodeForces - 659G

Fence Divercity

Time Limit: 2000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

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Description

Long ago, Vasily built a good fence at his country house. Vasily calls a fence good, if it is a series of n consecutively fastened vertical boards of centimeter width, the height of each in centimeters is a positive integer. The house owner remembers that the height of the i-th board to the left is h_i.

Today Vasily decided to change the design of the fence he had built, by cutting his top connected part so that the fence remained good. The cut part should consist of only the upper parts of the boards, while the adjacent parts must be interconnected (share a non-zero length before cutting out of the fence).

You, as Vasily's curious neighbor, will count the number of possible ways to cut exactly one part as is described above. Two ways to cut a part are called distinct, if for the remaining fences there is such i, that the height of the i-th boards vary.

As Vasily's fence can be very high and long, get the remainder after dividing the required number of ways by 1 000 000 007(10⁹ + 7).

Input

The first line contains integer n (1 ≤ n ≤ 1 000 000) — the number of boards in Vasily's fence.

The second line contains n space-separated numbers h₁, h₂, ..., h_n (1 ≤ h_i ≤ 10⁹), where h_i equals the height of the i-th board to the left.

Output

Print the remainder after dividing r by 1 000 000 007, where r is the number of ways to cut exactly one connected part so that the part consisted of the upper parts of the boards and the remaining fence was good.

Sample Input

Input

21 1

Output

0

Input

33 4 2

Output

13

Hint

From the fence from the first example it is impossible to cut exactly one piece so as the remaining fence was good.

All the possible variants of the resulting fence from the second sample look as follows (the grey shows the cut out part):

Source

Codeforces Round #346 (Div. 2)

//题意：

现在有n堵紧贴着的墙（墙是由砖块累成的），每堵墙都有一个高度，现在想将这n堵墙拆掉，并且拆墙有一定的规则，

1、每堵墙都不能把最下面的那块砖拆掉；

2、拆的砖头必须是连着的。

问一共有几种拆法？

//思路：

dp[i][0]存放前i列的所有的情况数；

dp[i][1]存放的是前i列全连通的情况数+第i+1列的情况数。

#include<stdio.h>#include<string.h>#include<math.h>#include<map>#include<queue>#include<stack>#include<algorithm>#include<iostream>#define INF 0x3f3f3f3f#define ull unsigned long long#define ll long long#define IN __int64#define N 1000010#define M 1000000007using namespace std;ll h[N],dp[N][2];int n;int main(){int i,j;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++)scanf("%lld",&h[i]),h[i]--;for(i=1;i<=n;i++){dp[i][0]=(dp[i-1][0]+dp[i-1][1]*min(h[i],h[i-1])+h[i])%M;dp[i][1]=(min(h[i+1],h[i])+min(min(h[i],h[i-1]),h[i+1])*dp[i-1][1])%M;}printf("%d\n",dp[n][0]);}return 0;}