ural 1247. Check a Sequence

1247. Check a Sequence

Time limit: 0.5 second
Memory limit: 64 MB

There is a sequence of integer numbers A1, A2, …, AS, and a positive integer N. It's known that all elements of the sequence {Ai} satisfy the restriction 0 ≤ Ai ≤ 100. Moreover, it's known that the sum of all elements of the sequence is equal to S + N. You are to write a program that given a sequence {Ai} and a number N will answer the question: is it true that for all 1 ≤ i ≤ j ≤ S the following inequality holds:

Ai + Ai+1 + … + Aj ≤ (j – i + 1) + N ?

Input

The first input line contains two separated with a space positive numbers S and N that do not exceed 30000. Then follow S lines with one number in a line that are elements of the sequence {Ai}.

Output

Output "YES", if the mentioned above inequality holds for all the values of the parameters i and j, and "NO" otherwise.

Samples

input	output
4 3 2 3 0 2	YES
4 5 1 0 5 3	NO

Problem Author: Alexander Mironenko
Problem Source: Ural State University Personal Programming Contest, March 1, 2003

Tags: none  (hide tags for unsolved problems)

Difficulty: 245

 

题意：问一个数列是否满足对于任意的i,j，都有sigma(a[k]) <= (j-i+1)+M, i<=k<=j，其中M给出，1<=n<=30000

分析：

其实我这个傻逼一开始以为是斜率dp。

后来发现，那个式子不就是

sigma(a[k]-1) <= M吗。。。

不就是对数列减一，然后做个最大子段和和M比较。。。

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 #include <iomanip>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
 #define Rep(i, t) for(int i = (0); i < (t); i++)
 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
 #define rep(i, x, t) for(int i = (x); i < (t); i++)
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define ft first
 #define sd second
 #define mk make_pair
 inline void SetIO(string Name)
 {
     string Input = Name+".in",
     Output = Name+".out";
     freopen(Input.c_str(), "r", stdin),
     freopen(Output.c_str(), "w", stdout);
 }

 inline int Getint()
 {
     int Ret = ;
     char Ch = ' ';
     bool Flag = ;
     while(!(Ch >= '' && Ch <= ''))
     {
         if(Ch == '-') Flag ^= ;
         Ch = getchar();
     }
     while(Ch >= '' && Ch <= '')
     {
         Ret = Ret *  + Ch - '';
         Ch = getchar();
     }
     return Flag ? -Ret : Ret;
 }

 const int N = ;
 int n, m, Arr[N], Ans;

 inline void Input()
 {
     scanf("%d%d", &n, &m);
     For(i, , n) scanf("%d", &Arr[i]);
 }

 inline void Solve()
 {
     int Cnt = Arr[] - ;
     Ans = Cnt;
     For(i, , n)
     {
         Cnt = max(Cnt + Arr[i] - , Arr[i] - );
         Ans = max(Ans, Cnt);
     }
     if(Ans <= m) puts("YES");
     else puts("NO");
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
     SetIO("F");
     #endif
     Input();
     Solve();
     return ;
 }