Kick Start Round D 2020(前两题)

https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff08?scoreboard_type=China

A.当前数比前面所有和后面一个都大的时候，算做破纪录一次。遍历一遍。

#include <bits/stdc++.h>

using namespace std;

vector<int> visiter(1000005,0);

int main()
{
    int T,N;
    int currmax=-1;
    int res=0;
    cin>>T;
    for(int i=0;i<T;i++){
        currmax=-1;
        res=0;
        cin>>N;
        for(int j=0;j<N;j++){
            cin>>visiter[j];
        }
        cout<<"Case #"<<i+1<<": ";
        if(N==1){
            cout<<1;
        }else{
            for(int j=0;j<N-1;j++){
                if(visiter[j]>currmax && visiter[j]>visiter[j+1]){
                    res++;
                }
                currmax=max(currmax,visiter[j]);
            }
            if(visiter[N-1]>currmax) res++;
            cout<<res;
        }
        cout<<endl;
    }


    return 0;
}

B.将多音高的音乐转换成只有4个音高的音乐，问最小破坏规则的次数。动态规划

#include <bits/stdc++.h>

using namespace std;

int notes[10005];
int dp[10005][4];

int main()
{
    int T,K;
    dp[0][0]=0;
    dp[0][1]=0;
    dp[0][2]=0;
    dp[0][3]=0;
    int res=0;
    cin>>T;
    for(int i=0;i<T;i++){
        cin>>K;
        for(int j=0;j<K;j++){
            cin>>notes[j];
        }
        cout<<"Case #"<<i+1<<": ";
        for(int j=1;j<K;j++){
            if(notes[j]>notes[j-1]){
                dp[j][0]=min(min(dp[j-1][0],dp[j-1][1]),min(dp[j-1][2],dp[j-1][3]))+1;
                dp[j][1]=min(min(dp[j-1][0],dp[j-1][1]+1),min(dp[j-1][2]+1,dp[j-1][3]+1));
                dp[j][2]=min(min(dp[j-1][0],dp[j-1][1]),min(dp[j-1][2]+1,dp[j-1][3]+1));
                dp[j][3]=min(min(dp[j-1][0],dp[j-1][1]),min(dp[j-1][2],dp[j-1][3]+1));
            }else if(notes[j]==notes[j-1]){
                dp[j][0]=min(min(dp[j-1][0],dp[j-1][1]+1),min(dp[j-1][2]+1,dp[j-1][3]+1));
                dp[j][1]=min(min(dp[j-1][0]+1,dp[j-1][1]),min(dp[j-1][2]+1,dp[j-1][3]+1));
                dp[j][2]=min(min(dp[j-1][0]+1,dp[j-1][1]+1),min(dp[j-1][2],dp[j-1][3]+1));
                dp[j][3]=min(min(dp[j-1][0]+1,dp[j-1][1]+1),min(dp[j-1][2]+1,dp[j-1][3]));
            }else if(notes[j]<notes[j-1]){
                dp[j][0]=min(min(dp[j-1][0]+1,dp[j-1][1]),min(dp[j-1][2],dp[j-1][3]));
                dp[j][1]=min(min(dp[j-1][0]+1,dp[j-1][1]+1),min(dp[j-1][2],dp[j-1][3]));
                dp[j][2]=min(min(dp[j-1][0]+1,dp[j-1][1]+1),min(dp[j-1][2]+1,dp[j-1][3]));
                dp[j][3]=min(min(dp[j-1][0],dp[j-1][1]),min(dp[j-1][2],dp[j-1][3]))+1;
            }
        }
        cout<<min(min(dp[K-1][0],dp[K-1][1]),min(dp[K-1][2],dp[K-1][3]));
        cout<<endl;
    }

    return 0;
}