Knot Puzzle——AT

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 Copy

3 50
30 20 10

样例输出 Copy

Possible

提示

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

题意，将相邻的两端绳子打上结拴在一起，问给出某长度能否将某个结打开，题目较水

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
///#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
///char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll mod=1e9+7;
const ll inf=0x3f3f3f3f;
const int maxn=1e6+9;
ll cnt=0;
///int zhan[maxn];
int num[maxn];
int top;
int main()
{
    ll n=read,l=read;
    vector<ll>vet(n);
    for(int i=0;i<n;i++) vet[i]=read;
    for(int i=0;i<n-1;i++)
    {
        if(vet[i]+vet[i+1]>=l){
            printf("Possible
");
            return 0;
        }
    }
    printf("Impossible
");
    return 0;
}