hdu 3555 Bomb(数位DP,4级)
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 4443 Accepted Submission(s): 1538
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?
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.
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 15HintFrom 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.
Author
fatboy_cw@WHU
Source
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zhouzeyong
很水的数位DP
#include<cstdio> #include<cstring> #include<iostream> #define FOR(i,a,b) for(int i=a;i<=b;++i) #define clr(f,z) memset(f,z,sizeof(f)) #define LL __int64 using namespace std; LL dp[40][10][2];//位数,最后一个数,是否有49 int bit[40],pos; LL DP(int pp,int ee,bool has,bool big) { if(pp==0)return has; if(big&&dp[pp][ee][has]!=-1)return dp[pp][ee][has]; int kn=big?9:bit[pp]; LL ret=0; FOR(i,0,kn) { ret+=DP(pp-1,i,has||(ee==4&&i==9),big||(i!=kn)); } if(big)dp[pp][ee][has]=ret; return ret; } LL get(LL x) { pos=0; while(x) { bit[++pos]=x%10;x/=10; } clr(dp,-1); return DP(pos,0,0,0); } int main() { int cas; while(cin>>cas) { while(cas--) { LL n; cin>>n; cout<<get(n)<<endl; } } }
The article write by nealgavin