hdu 3555 Bomb-优快云博客

本文链接：https://blog.youkuaiyun.com/libing923/article/details/8864106

本文探讨了一个计数问题，即在从1到N的整数范围内，有多少个数包含子序列“49”。通过动态规划的方法给出了详细的解决方案，并提供了一段C++代码实现。

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Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 3299 Accepted Submission(s): 1168

Problem Description

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

HintFrom 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.

题目大意：求区间[1,N]内有多少个包含"49"的数

dp[i][0] 遍历到第i位未出现过"49"且前一个不是4;

dp[i][1] 遍历到第i位未出现过"49"且前一个是4;

dp[i][2] 遍历到第i位出现过"49";

#include<cstdio>
#include<cstring>
#define ll __int64
ll dp[20][3];
int a[20];
int new_s(int J,int d){
	if(J==0)return d==4?1:0;
	if(J==1){
		if(d==4)return 1;
		if(d==9)return 2;
		return 0;
	}
	return 2;
}
ll dfs(int i,int J,bool e){
	if(i==-1)return J==2;
	if(!e&&dp[i][J]!=-1)return dp[i][J];
	ll res=0;
	int u=e?a[i]:9,d;
	for(d=0;d<=u;d++)res+=dfs(i-1,new_s(J,d),e&&d==u);
	return e?res:dp[i][J]=res;
}
ll cal(ll n){
	int i=0;
	while(n)a[i++]=n%10,n/=10;
	return dfs(i-1,0,1);
}
int main(){
	int T;
	memset(dp,-1,sizeof(dp));
	ll n;
	scanf("%d",&T);
	while(T--)scanf("%I64d",&n),printf("%I64d\n",cal(n));
	return 0;
}