ural 1247. Check a Sequence

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1247. Check a SequenceTime limit: 0.5 second
Memory limit: 64 MBThere 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 ?InputThe 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}.OutputOutput "YES", if the mentioned above inequality holds for all the values of the parameters i and j, and "NO" otherwise.Samples

input	output
4 32302	YES
4 510 53	NO

Problem Author: Alexander Mironenko
Problem Source: Ural State University Personal Programming Contest, March 1, 2003Tags: none  ()Difficulty: 245 題意：問一個數列是否滿足對於任意的i,j，都有sigma(a[k]) <= (j-i+1)+M, i<=k<=j，其中M給出，1<=n<=30000分析：其實我這個傻逼一開始以為是斜率dp。後來發現，那個式子不就是sigma(a[k]-1) <= M嗎。。。不就是對數列減一，然後做個最大子段和和M比較。。。

 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_front28 #define pub push_back29 #define pof pop_front30 #define pob pop_back31 #define ft first32 #define sd second33 #define mk make_pair34 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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ural 1247. Check a Sequence