Codeforces Gym 100372A

很少有人做Gym的题目，但其实Gym的题目还是不错的，先来水题一道 100372A

Sergey and reduction

Input file: stdin
Output file: stdout
Time limit: 2 sec
Memory limit: 256 Мb
Sergey has an array of n of integers a1, a2, · · · , an. He wants to be able to respond to two types of queries:
• 0 l r e — for each i (l ≤ i ≤ r) execute ai = ai − e
• 1 l r — find out the count of i (l ≤ i ≤ r), that ai ≤ 0
Help Sergey to make an answer to m queries.
Input
The first line contains two integers n, m (2 ≤ n ≤ 4000, 1 ≤ m ≤ 35000). The next line contains n integers a1, a2,
· · · , an (0 ≤ ai ≤ 109
). The next m lines contains queries, as was discovered in legend. If the query is like 0 l r e,
works the following restriction 1 ≤ l ≤ r ≤ n, 0 ≤ e ≤ 109
.
Output
For each query like 1 l r print an answer in a separate line.
Example
stdin stdout
3 6
1 2 3
0 2 3 1
1 1 3
0 2 2 1
1 1 3
0 1 3 1
1 1 2
0
1
2

题目大意如下 ： 有0号操作和1号操作 {
0号操作 ：将A[l] ~ A[r] 中的数值全部减去e
1号操作 ：在A[l]~ A[r] 中查找数值小于0的个数}

#include <iostream>#include <cstdio> #include <cmath>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <string>#include <cstring>using namespace std;const int maxn = 40000;int q[40001];void input0(int l,int r,int e){    for (int i=l;i<=r;++i){        q[i] -= e;    }    return;}int input1(int l,int r){    int cnt=0;    for (int i=l;i<=r;++i){        cnt += (q[i]<=0);    }    return cnt;}int main(){    int n,m;    cin >> n >> m;    for (int i=1;i<=n;++i)cin >> q[i];    for (int i=1;i<=m;++i){        int t;        scanf("%d",&t);        if (!t){            int l,r,e;            scanf("%d%d%d",&l,&r,&e);            input0(l,r,e);        }        else{            int l,r;            scanf("%d%d",&l,&r);            printf("%d\n",input1(l,r));        }    }    return 0;}