中油5583: Knot Puzzle

5583: Knot Puzzle

时间限制: 2 Sec   内存限制: 256 MB
提交: 232   解决: 80
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题目描述

We have N pieces of ropes, numbered 1 through N. The length of piece i is ai.

At first, for each i(1≤i≤N−1), piece i and piece i+1 are tied at the ends, forming one long rope with N−1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly:

Choose a (connected) rope with a total length of at least L, then untie one of its knots.
Is it possible to untie all of the N−1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.

Constraints
2≤N≤105
1≤L≤109
1≤ai≤109
All input values are integers.

输入

The input is given from Standard Input in the following format:

N L
a1 a2 … an

输出

If it is not possible to untie all of the N−1 knots, print Impossible.

If it is possible to untie all of the knots, print Possible .

样例输入

3 50
30 20 10

样例输出

Possible

提示

If the knot 1 is untied first, the knot 2 will become impossible to untie.

来源

AtCoder Grand Contest 002 

//思路：思考一下什么时候是一定能解开的
//1.有一段绳子>l的时候，将这段绳子留在最后，一定能全解开
//2.根据1，可类比得出当存在两段相邻绳子的长度和>=l，也一定能解开
//3.所以反过来可以推出，如果所有两段相邻绳子的长度和<l，一定解不开

#include<bits/stdc++.h>
using namespace std;
int s[100005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int l;
        scanf("%d",&l);
        for(int i=0; i<n; i++)
        {
            scanf("%d",&s[i]);
        }
        int num = 0;
        for(int i=0; i<n-1; i++)
        {
            if(s[i]+s[i+1]<l)
            {
                num++;
            }
        }
        if(num == n-1) printf("Impossible\n");
        else printf("Possible\n");
    }
}