【1】【数位DP】HDU3555 Bomb
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Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 14224 Accepted Submission(s): 5124
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
3
1
50
500
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.
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;__int64 dp[30][3];__int64 bit[30];__int64 dfs(__int64 pos,__int64 have,__int64 limit){ __int64 have_x,ans=0; if(pos<=0)return have==2; if(!limit&&dp[pos][have]!=-1)return dp[pos][have]; __int64 num=limit?bit[pos]:9; for(__int64 i=0;i<=num;i++){ have_x=have; if(have==0&&i==4)have_x=1; if(have==1&&i!=4)have_x=0; if(have==1&&i==9)have_x=2; ans+=dfs(pos-1,have_x,limit&&i==num); } if(!limit)dp[pos][have]=ans; return ans;}__int64 solve(__int64 n){ memset(bit,0,sizeof(bit)); memset(dp,-1,sizeof(dp)); __int64 len=0; while(n) { bit[++len]=n%10; n/=10; } return dfs(len,0,1);}int main(){ __int64 t; scanf("%I64d",&t); while(t--) { __int64 n; scanf("%I64d",&n); printf("%I64d\n",solve(n)); } return 0;}
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