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

inputoutput
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
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。。。
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <deque>
 6 #include <vector>
 7 #include <queue>
 8 #include <iostream>
 9 #include <algorithm>
10 #include <map>
11 #include <set>
12 #include <ctime>
13 #include <iomanip>
14 using namespace std;
15 typedef long long LL;
16 typedef double DB;
17 #define For(i, s, t) for(int i = (s); i <= (t); i++)
18 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
19 #define Rep(i, t) for(int i = (0); i < (t); i++)
20 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
21 #define rep(i, x, t) for(int i = (x); i < (t); i++)
22 #define MIT (2147483647)
23 #define INF (1000000001)
24 #define MLL (1000000000000000001LL)
25 #define sz(x) ((int) (x).size())
26 #define clr(x, y) memset(x, y, sizeof(x))
27 #define puf push_front
28 #define pub push_back
29 #define pof pop_front
30 #define pob pop_back
31 #define ft first
32 #define sd second
33 #define mk make_pair
34 inline void SetIO(string Name)
35 {
36     string Input = Name+".in",
37     Output = Name+".out";
38     freopen(Input.c_str(), "r", stdin),
39     freopen(Output.c_str(), "w", stdout);
40 }
41 
42 inline int Getint()
43 {
44     int Ret = 0;
45     char Ch = ' ';
46     bool Flag = 0;
47     while(!(Ch >= '0' && Ch <= '9'))
48     {
49         if(Ch == '-') Flag ^= 1;
50         Ch = getchar();
51     }
52     while(Ch >= '0' && Ch <= '9')
53     {
54         Ret = Ret * 10 + Ch - '0';
55         Ch = getchar();
56     }
57     return Flag ? -Ret : Ret;
58 }
59 
60 const int N = 30010;
61 int n, m, Arr[N], Ans;
62 
63 inline void Input() 
64 {
65     scanf("%d%d", &n, &m);
66     For(i, 1, n) scanf("%d", &Arr[i]);
67 }
68 
69 inline void Solve()
70 {
71     int Cnt = Arr[1] - 1;
72     Ans = Cnt;
73     For(i, 2, n)
74     {
75         Cnt = max(Cnt + Arr[i] - 1, Arr[i] - 1);
76         Ans = max(Ans, Cnt);
77     }
78     if(Ans <= m) puts("YES");
79     else puts("NO");
80 }
81 
82 int main()
83 {
84     #ifndef ONLINE_JUDGE 
85     SetIO("F");
86     #endif 
87     Input();
88     Solve();
89     return 0;
90 }
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posted @ 2015-11-14 16:12  yanzx6  阅读(186)  评论(0)  编辑  收藏  举报