Bomb （数位dp）_牛客博客

The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

Output

For each test case, output an integer indicating the final points of the power.

Sample Input

Sample Output

0
1
15

Hint

From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499",
so the answer is 15.

求1~n中含有“49”的数的个数

思路1：前面“不要62”那道题知道了怎么求0~n不含62的数的个数怎么求，改一改求不含49的数的个数，再用n+1-solve(n)

注意：solve(n)求的是0~n中不含49的个数

注意这道题的数据范围，long long

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int  a[30];
ll dp[30][2];
ll dfs(int pos,int pre,int sta,bool limit){
	if(pos==0) return 1;
	if(!limit&&dp[pos][sta]!=-1) return dp[pos][sta];
	int up=limit?a[pos]:9;
	ll tmp=0;
	for(int i=0;i<=up;i++){
		if(pre==4&&i==9) continue;
		tmp+=dfs(pos-1,i,i==4,limit&&i==a[pos]);
	} 
	if(!limit) dp[pos][sta]=tmp;
	return tmp; 
}
ll solve(ll x){
	int pos=0;
	while(x){
		a[++pos]=x%10;
		x/=10;
	}
	return dfs(pos,-1,0,true);
}
int main(){
	int T;
	scanf("%d",&T);	
	while(T--){
		ll n;
		scanf("%lld",&n);	
		memset(dp,-1,sizeof(dp));
		printf("%lld\n",n+1-solve(n));
	}
	return 0;
}

思路2：直接求含49的数的个数

dp[i][0]:长度为i但是不包含49的方案数
dp[i][1]:长度为i且不含49但是最后一位是4的方案数
dp[i][2]:长度为i且包含49的方案数

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int  a[30];
ll dp[30][3];
ll dfs(int pos,int sta,bool limit){ 
	if(pos==0) return sta==2;
	if(!limit&&dp[pos][sta]!=-1) return dp[pos][sta];
	int up=limit?a[pos]:9;
	ll tmp=0;
	for(int i=0;i<=up;i++){
		if(sta==2||(sta==1&&i==9)) 
		tmp+=dfs(pos-1,2,limit&&i==a[pos]);
		else if(i==4)
		tmp+=dfs(pos-1,1,limit&&i==a[pos]);
		else
		tmp+=dfs(pos-1,0,limit&&i==a[pos]);
	} 
	if(!limit) dp[pos][sta]=tmp;
	return tmp; 
}
ll solve(ll x){
	int pos=0;
	while(x){
		a[++pos]=x%10;
		x/=10;
	}
	return dfs(pos,0,true);
}
int main(){
	int T;
	scanf("%d",&T);	
	while(T--){
		ll n;
		scanf("%lld",&n);	
		memset(dp,-1,sizeof(dp));
		printf("%lld\n",solve(n));
	}
	return 0;
}