hdu 3555 Bomb(数位DP,4级)
2013-08-19 11:41
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Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 4443 Accepted Submission(s): 1538
[align=left]Problem Description[/align]
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?
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case, output an integer indicating the final points of the power.
[align=left]Sample Input[/align]
3 1 50 500
[align=left]Sample Output[/align]
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.
[align=left]Author[/align]
fatboy_cw@WHU
[align=left]Source[/align]
2010 ACM-ICPC Multi-University
Training Contest(12)——Host by WHU
[align=left]Recommend[/align]
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; } } }
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