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P1037 production (high precision & DFS)

2021-08-10 08:04:59 wx6110fa547fd20

P1037 Generating numbers ( Hyperfine &DFS)

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Given as a large integer ,k The conversion of middle numbers . Find out how many different numbers can be converted into .

Ideas : utilize DFS Find every bit that can be converted , According to the principle of multiplication, multiplying the situation of each bit is the answer .

#include<bits/stdc++.h>
using namespace std;
int k,ans[100]={1},vis[10],tmp;//vis[i] Mark whether the number has been accessed  
pair<int,int>p[20];
void dfs(int x){ // Search every possible situation  
	for(int i=1;i<=k;i++)
		if(!vis[p[i].second]&&p[i].first==x)
			vis[p[i].second]=1,tmp++,dfs(p[i].second);
}
void jc(int *a,int b){ // High precision multiplication and low precision  ( Analog high-precision factorial ) 
	 for(int i=0,p=0;i<100;i++)
	{
		 a[i]=a[i]*b+p;
		 p=a[i]/10;
		 a[i]%=10;
	}
}
int main(){
	string a;
	cin>>a>>k;
	for(int i=1;i<=k;i++) cin>>p[i].first>>p[i].second;
	for(int i=0;i<a.size();i++){
		int num=a[i]-'0';
		memset(vis,0,sizeof vis);
		vis[num]=1,tmp=1;
		dfs(num);
		jc(ans,tmp);// Multiplication principle  
	}
	for(int i=99,f=0;i>=0;i--)
	{
		if(ans[i]) f=1;
		if(f) printf("%d",ans[i]);
	}
	puts("");
	return 0;
}

      
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