codeforces740B

本文介绍CodeForces-740B题目“Alyona and Flowers”的解题思路与实现方法,该题涉及子数组选择,旨在最大化主人公Alyona的幸福感。通过分析花束的情绪值,采用简单的遍历策略快速得出最优解。

Alyona and flowers

 CodeForces - 740B 

Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.

Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.

For example, consider the case when the mother has 5 flowers, and their moods are equal to 1,  - 2, 1, 3,  - 4. Suppose the mother suggested subarrays (1,  - 2), (3,  - 4), (1, 3), (1,  - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:

  • the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
  • the second flower adds ( - 2)·1 =  - 2, because he is in one of chosen subarrays,
  • the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
  • the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
  • the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.

Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!

Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.

The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).

The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].

Each subarray can encounter more than once.

Output

Print single integer — the maximum possible value added to the Alyona's happiness.

Examples

Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0

Note

The first example is the situation described in the statements.

In the second example Alyona should choose all subarrays.

The third example has answer 0 because Alyona can choose none of the subarrays.

 

sol:这题意TMD居然看成了一道数据结构题(智减inf)
搞出前缀和,显然如果是正数就算入答案,否则不算
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0');    return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=105;
int n,m,a[N],Qzh[N];
int S[N];
int main()
{
    int i,j,ans=0;
    R(n); R(m);
    for(i=1;i<=n;i++) R(a[i]);
    for(i=1;i<=n;i++) Qzh[i]=Qzh[i-1]+a[i];
    for(i=1;i<=m;i++)
    {
        int l=read(),r=read();
        if(Qzh[r]-Qzh[l-1]>0) ans+=(Qzh[r]-Qzh[l-1]);
    }
    Wl(ans);
    return 0;
}
/*
input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
output
7

input
4 3
1 2 3 4
1 3
2 4
1 1
output
16

input
2 2
-1 -2
1 1
1 2
output
0
*/
View Code

 

 

转载于:https://www.cnblogs.com/gaojunonly1/p/10638740.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值