title: ACM-刷题记录-14届NEFU校赛
tags:
- ACM
abbrlink: 77f448c2
date: 2020-12-03 16:34:50
P2031凯撒密码
#include<bits/stdc++.h>using namespace std;int main(){string s;int d;while(cin>>s){cin>>d;int len=s.length();for(int i=0;i<len;i++){s[i]=s[i]-'a'-d;if(s[i]<0) s[i]=26+s[i];s[i]=s[i]%26+'a';printf("%c", s[i]);}cout<<endl;}return 0;}
P2033神秘数字
#include<bits/stdc++.h>using namespace std;int main(){long long n,m,ans=0;cin>>n>>m;//cout<<pow(10,n)<<endl<<pow(10,n+1)<<endl;for(int i=pow(10,n-1);i<pow(10,n);i++){int num=i,sum=0;while(num){sum+=num%10;num/=10;}if(sum==m){//cout<<i<<endl;ans+=i;}}cout<<ans<<endl;return 0;}
P100 快来找一找
#include<iostream>#include<cstdio>using namespace std;int main(){int m,maxnum=0;while(cin>>m){int tmp;for(int i=0;i<m;i++){cin>>tmp;maxnum=max(maxnum,tmp);}cout<<maxnum<<endl;maxnum=0;}}
P2034 第几个质数
老夫还是不用筛
#include<bits/stdc++.h>using namespace std;int prime[1000+5];int isprime(int x){if(prime[x]!=-1) return prime[x];for(int i=2;i<=sqrt(x);i++){if(x%i==0) return prime[x]=0;}return prime[x]=1;}int main(){int n,number=1;memset(prime, -1, sizeof(prime));prime[1]=0; prime[2]=1;while(cin>>n){if(!isprime(n)){cout<<"NO"<<endl;continue;}for(int i=1;i<n;i++){if(isprime(i)){number++;}}cout<<number<<endl;number=1;}return 0;}
P2039 why的概率论
#include<bits/stdc++.h>using namespace std;int main(){int t,x,y;cin>>t;while(t--){cin>>x>>y;if(!x) cout<<"0 / 1"<<endl;else{y+=x;x*=4;cout<<x/__gcd(x,y)<<" / "<<y/__gcd(x,y)<<endl;}}return 0;}
P2065 小x爱料理
#include<bits/stdc++.h>using namespace std;int main(){int t;cin>>t;while(t--){int u,v,w,a,b,c;int n1,n2,n3,ans;cin>>u>>v>>w>>a>>b>>c;n1=u/a;n2=v/b;n3=w/c;ans=min(n1, min(n2, n3));cout<<ans<<endl;}return 0;}
库特的鸽鸽们
#include<bits/stdc++.h>using namespace std;const int MAXN=2e5+10;int n,q,a[MAXN],c[MAXN],p[MAXN],maxn[MAXN];int main(){cin>>n;for(int i=1;i<=n;i++){cin>>a[i];}cin>>q;for(int i=1;i<=q;i++){int op,x,y;cin>>op;if(op==1){cin>>x>>y;c[x]=y;p[x]=i;}else cin>>maxn[i];}for(int i=q-1;i>=1;i--){maxn[i]=max(maxn[i], maxn[i+1]);}for(int i=1;i<=n;i++){if(p[i]) cout<<max(c[i], maxn[p[i]])<<" ";else cout<<max(maxn[1], a[i])<<" ";}//if(p[n]) cout<<max(c[n], maxn[p[n]]);//else cout<<max(maxn[1], a[n]);return 0;}
