HDU3555 Bomb —— 数位DP
2024-08-29 07:26:10
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3555
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 18739 Accepted Submission(s): 6929
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?
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
3
1
50
500
1
50
500
Sample Output
0
1
15
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.
题解:
数位DP通用:dp[pos][sta1][sta2][……]
表示:当前位为pos,之前的状态为sta1*sta2*……stan。n为限制条件的个数。
回到此题,限制条件有两个: 1.上一位是否为4; 2.之前是否已经出现49。
类似题目:http://blog.csdn.net/dolfamingo/article/details/72848001
代码如下:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+;
const int maxn = +; LL n;
LL a[maxn], dp[maxn][]; LL dfs(int pos, int status, bool lim)
{
if(!pos) return status==;
if(!lim && dp[pos][status]!=-) return dp[pos][status]; LL ret = ;
int m = lim?a[pos]:;
for(int i = ; i<=m; i++)
{
int tmp_status;
if(status== || status== && i==)
tmp_status = ;
else if(i==)
tmp_status = ;
else
tmp_status = ; ret += dfs(pos-, tmp_status, lim&&(i==m));
} if(!lim) dp[pos][status] = ret;
return ret;
} int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%lld",&n);
int p = ;
while(n)
{
a[++p] = n%;
n /= ;
}
memset(dp,-, sizeof(dp));
LL ans = dfs(p, , );
printf("%lld\n",ans);
}
}
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