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CF 599 div2 A - A pile of stones

作者:互联网

题意: 给出n个+ - 操作符  求出结束的最小值(当然 过程中不能减到小于0) 

 

求出遍历完成后的总的变化值  和遍历过程中的最小值  这个值就是最小的初始值(显然 题目可以转化为求最小初始值)

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define pb push_back
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
#define lson l,m,pos<<1
#define rson m+1,r,pos<<1|1
const int N=1000+5;

int main()
{
    int n;
    RI(n);
    char ch;
    int ans=0;
    int minn=0;
    while(n--)
    {
        cin>>ch;
        if(ch=='+')ans++;
        else ans--;
        minn=min(minn,ans);
    }
    cout<<-minn+ans;


    return 0;
}
View Code

 

 

 

 

官方题解:

反过来遍历 更加巧妙

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define pb push_back
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
#define lson l,m,pos<<1
#define rson m+1,r,pos<<1|1
const int N=400000+5;

int main()
{
    int n;
    RI(n);
    string s;
    cin>>s;
    int ans=0;
    int m=0;
    repp(i,n-1,0)
    {
        if(s[i]=='+')
            m++;
        else m--;
        ans=max(ans,m);
    }
    cout<<ans;


    return 0;
}
View Code

 

标签:stones,599,scanf,d%,long,int,pile,ans,define
来源: https://www.cnblogs.com/bxd123/p/10856892.html