牛客-西南科技大学第十六届ACM程序设计竞赛暨绵阳市邀请赛

A-找规律

2*9-13=5，映射13次一次循环

/*
* problem:找规律
* method:映射循环
* date:2020/07/07
*/
#include<iostream>
#include<string>
#define LL long long
using namespace std;
string s1[14],s2[14];
int f[14],n=13;
int main() {
	int i,j;
	while(cin>>s1[0]) {
		for(i=1; i<n; i++) cin>>s1[i];
		for(i=0; i<n; i++) cin>>s2[i];
		for(i=0; i<n; i++) {
			for(j=0; j<n; j++) {
				if(s1[i]==s2[j]) {
					f[i]=j;
					break;
				}
			}
		}
		for(i=0; i<n; i++) {
			int t=i;
			for(j=0; j<9; j++) t=f[t];
			s2[t]=s1[i];
		}
		for(i=0; i<n-1; i++) {
			cout<<s2[i]<<' ';
		}
		cout<<s2[n-1]<<endl;
	}
	return 0;
}

B-签到题

找公式，(n-1)!*(n-1)*n/2

/*
* problem:签到题
* method:阶乘、取余、空间换时间
* date:2020/07/07
*/
#include<iostream>
using namespace std;
long long int MAX=1e9+7;
long long int f[100005];
int main(){
    long long n;
    f[0]=1;
    for(int i=1;i<100005;i++){
    	f[i]=(i*f[i-1])%MAX;
	}
    while(cin>>n){
        cout<<(((n*(n-1)/2%MAX)*f[n])%MAX)<<endl;
    }
    return 0;
}

C-救救AR

找规律：两个A，n-2个R，cell(n/2)+floor(n/2)=n

/*
* problem:救救AR
* method:找规律、string
* date:2020/07/07
*/
#include<iostream>
#include<string>
#define LL long long
using namespace std;
int main(){
	int n;
	string s;
	while(cin>>n){
		if(n<4){
			cout<<"-1"<<endl;
		}else{
			s="";
			int i,x=n/2,y=n-x;
			s+="A";
			for(i=0;i<y-x;i++) s+="R";
			s+="A";
			for(i=0;i<x;i++) s+="R";
			cout<<s<<endl;
		}
		
	}
	return 0;
}

D-ar采蘑菇

动态规划，位表示集合

/*
* problem:ar采蘑菇
* method:dp、位压缩表示集合
* date:2020/07/11
*/
#include<iostream>
#include<string>
#include<utility>
#define LL long long
using namespace std;
typedef pair<int,int> P;
int n,m,k,t;
P p[5];
int dp[105][105];
int popcount(int x) {
	int count=0;
	while(x) {
		x=(x&(x-1));
		count++;
	}
	return count;
}
void bfs(int x,int y,int z) {
	int i,j,num,nx,ny,nz;
	if(x==n&&y==m) return;
	for(i=0; i<k; i++) {
		nx=x+p[i].first;
		ny=y+p[i].second;
		nz=z|(1<<i);
		num=popcount(nz);
		if(nx<=n&&ny<=m&&num>dp[ny][nx]){
			dp[ny][nx]=num;
			bfs(nx,ny,nz);
		} 
	}
}
int main() {
	int i,j,num;
	cin>>t;
	while(t--) {
		num=0;
		cin>>n>>m>>k;
		for(i=0;i<=m;i++){
			for(j=0;j<=n;j++) dp[i][j]=0;
		}
		for(i=0; i<k; i++) {
			string s;
			p[i].first=0;
			p[i].second=0;
			cin>>s;
			for(j=0; j<s.size(); j++) {
				if(s[j]=='R') {
					p[i].first++;
				} else if(s[j]=='U') {
					p[i].second++;
				}
			}
		}
		bfs(0,0,0);
		cout<<dp[m][n]<<endl;
	}
	return 0;
}

E-呼兰河传

筛素数法、快速幂、LCM

/*
* problem:
* method:
* date:
*/
#include<iostream>
#include<cmath>
#define ll long long
using namespace std;
const int N=1e6+5,MAX=1e5+5,MOD=1e9+9;
int prime[N],p[N],power[N],mp[N],cnt;
int max(int a,int b){
	return a>b?a:b;
}
void init(){//筛素数
	prime[0]=1;
	prime[1]=1;
	for(int i=2;i<=sqrt(MAX);i++){
		if(!prime[i]){
			p[cnt++]=i;
			mp[i]=cnt-1;
			for(int j=2;i*j<=MAX;j++) prime[i*j]=1;
		}
	}
}
ll fastPower(ll base,ll power){
	ll result=1;
	while(power>0){
		if(power&1){
			result=result*base%MOD;
			power--;
		}
		base=base*base%MOD;
		power>>=1;
	}
	return result;
}
int main(){
	int n,i,j,x;
	init();
	cin>>n;
	while(n--){
		cin>>x;
		for(i=0;i<cnt&&x>=p[i];i++){
			int cur=0;
			if(x%p[i]==0){
				while(x%p[i]==0){
					cur++;
					x/=p[i];
				}
			}
			power[i]=max(power[i],cur);
		}
		if(x>1){
			int j=0;
			if(!mp[x]){
				p[cnt++]=x;
				mp[x]=cnt-1;
				j=cnt-1;
			}else{
				j=mp[x];
			}
			power[j]=max(1,power[j]);
		}
	}
	ll lcm=1;
	for(i=0;i<cnt;i++){
		if(power[i]) lcm=lcm*fastPower(p[i],power[i])%MOD;
	}
	cout<<lcm<<endl;
	return 0;
}