2019年湘潭大学程序设计竞赛（重现赛）

Posted 2021-11-27 yjing814

2019年湘潭大学程序设计竞赛（重现赛） 

A:Who‘s better?

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n1,p1,s1,n2,p2,s2;
    cin>>n1>>p1>>s1;
    cin>>n2>>p2>>s2;
    if(n1!=n2){
        if(n1>n2)cout<<"1\n";
        else cout<<"2\n";
        return 0;
    }
    if(p1!=p2){
        if(p1<p2)cout<<"1\n";
        else cout<<"2\n";
        return 0;
    }
    if(s1!=s2){
        if(s1<s2)cout<<"1\n";
        else cout<<"2\n";
        return 0;
    }
    cout<<"God\n";
    return 0;
}

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B:Number

#include<bits/stdc++.h>
using namespace std;
int t,n;
int main(){
    scanf("%d",&t);
    while(t--){
        int ans=0;
        scanf("%d",&n);
        while(n>1){
            int p=n%10;
            if(p==0){
                ans++;
                n/=10;
            }
            else{
                n=n+10-p;
                ans=ans+10-p;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

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C:Math Problem

（a*k+1）³=a³*k³+3a²k²+3ak+1=(a²*k³+3a¹k²+3k)a+1=a*x+1;

即（a*k+1）的3次方也是a的倍数+1；

#include<bits/stdc++.h>
using namespace std;
int t,l,r;

int main(){
   
    
    scanf("%d",&t);
    while(t--){
        long long ans=0;
        scanf("%d%d",&l,&r);
        int p=l/192,q=r/192;
        if(l%192>1)p++;
        if(r%192<1)q--;
        for(int i=p;i<=q;i++){
            ans=ans+(long long )i*192+1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

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D:Stone

每次取最大堆与任一相邻合并，结果 ： 总和-max

#include<bits/stdc++.h>
using namespace std;
const int INF=1e9+10;
const int N=1e5;
int t,n;
long long a[N];
 
int main(){
    scanf("%d",&t);
    while(t--){
        long long ans=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%lld",&a[i]);
        }
        sort(a,a+n);
        for(int i=0;i<n-1;i++){
            ans+=(long long )a[i];
        }
         
        printf("%lld\n",ans);
    }
    return 0;
}

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F:Black & White

二分+前缀和

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+100;
int t,n,m;
char str[N];
int p[N];
int solve(int x){
    for(int i=0;i+x<=n;i++){
        if(p[i+x]-p[i]+m>=x||(i+x-p[i+x])-(i-p[i])+m>=x) return 1;
    }
    return 0;
}
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%s",&n,&m,str);
        memset(p,0,sizeof(p));
        for(int i=1;i<=n;i++){
            if(str[i-1]==‘1‘){
                p[i]=p[i-1]+1;
            }
            else{
                p[i]=p[i-1];
            }
        }
        int l=0,r=n,mid;
        while(l<=r){
            mid=(l+r)/2;
            if(solve(mid)) l=mid+1;
            else r=mid-1;
        }
        printf("%d\n",r);
    }
    return 0;
}

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