4.22代码

4.22

1. codeforces #636（Div.3）C题：简单思维

题目大意：给一个序列，求最长交替子序列（先后顺序不变，正负交替）的最大和。

想法

输入的过程判断，flag记录前一个数的正负，pre记录前一个值；如果正负交替就更新sum，如果正负相同，sum-=，+=，记最大值。

出现问题：超时

修正：别人的代码；修改之后也超时，但是read()就不超时——我的代码read还是超时…

//超时代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define inf 0x3f3f3f3f

using namespace std;
typedef long long ll;
const int maxn = 2e5+10;
ll read(){ll x; if(scanf("%lld",&x)==-1)exit(0); return x;}
int vis[maxn], dp[maxn], cntt[maxn];


//超时
//read()还是超时
int main(){
    int t;
    ll n, cnt = 0;
    cin>>t;
    ll flag = 0, pre = 0;
    while(t--){
        memset(dp, 0, sizeof(dp));
        memset(vis, 0, sizeof(vis));
        memset(cntt, 0, sizeof(cntt));
        n = read();
        cnt = 0;
        ll sum = 0;
        for(int i = 1; i <= n; i++){
            vis[i] = read();
            //cout<<1<<" "<<vis[i]<<endl;
            if(i == 1){
                cnt++;
                pre= flag = vis[i];
                cntt[i] = cnt;
                sum+=vis[i];
                //cout<<2<<" "<<vis[i]<<endl;
            }
            else if(flag*vis[i] < 0){
                pre = flag = vis[i];
                cnt++;
                cntt[i] = cnt;
                sum+=vis[i];
                //cout<<3<<" "<<vis[i]<<endl;
            }
            else if(vis[i]>pre){
                    sum-=pre;
                    sum+=vis[i];
                    pre = vis[i];
                    cntt[i] = cnt;

                    //cout<<4<<" "<<vis[i]<<endl;
            }
        }
        /*
        for(int i = n; i >= 1; i--){
            if(cntt[i] == cnt){
                sum+=vis[i];
                cnt--;
            }
        }
        */

        printf("%lld\n", sum);
    }
    return 0;
}

思路：read()读入之后，依次遍历处理序列，m找到连续序列中最大值，等到正负交替（即序列交替时），sum+=m。

//AC代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define inf 0x3f3f3f3f

using namespace std;
typedef long long ll;
const int maxn = 2e5+10;
ll read(){ll x; if(scanf("%lld",&x)==-1)exit(0); return x;}
int vis[maxn], dp[maxn], cntt[maxn];

//read()，解决超时问题

int main(){
    int t;
    ll n, cnt = 0;
    cin>>t;
    ll flag = 0, pre = 0;
    while(t--){

        n = read();

        for(int i = 1; i <= n; i++){
            vis[i] = read();

        }
        ll sum = 0;
        int m = inf*(-1);
        flag = abs(vis[1])/vis[1];
        for(int i = 1; i <= n; i++){
            if(vis[i]*flag>0){
                m = max(m, vis[i]);
            }
            else{
                flag = abs(vis[i])/vis[i];
                sum+=m;
                //cout<<"          "<<m<<endl;
                m = vis[i];
            }
        }

        sum+=m;
        printf("%lld\n", sum);
    }
    return 0;
}