HDU3555 Bomb
Posted Forever_goodboy
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Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Problem DescriptionThe 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?InputThe 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.OutputFor each test case, output an integer indicating the final points of the power.Sample Input3 1 50 500Sample 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.
Solution:
很裸的数位dp,记忆化搜索好啊,不过数组开小了,炸了几次。
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 using namespace std; 6 long long T,dp[105][15],bit[105]; 7 #define int long long 8 long long n; 9 long long dfs(int pos,int pre,bool Limit) { 10 if(pos<=0&&pre>=0) return 1; 11 if(pos<=0) return 0; 12 if(pre>=0&&!Limit&&dp[pos][pre]!=-1) return dp[pos][pre]; 13 int End=Limit?bit[pos]:9; 14 long long ans=0; 15 for(int now,i=0; i<=End; ++i) { 16 if(pre==4&&i==9) continue; 17 now=i; 18 if(pre<0&&!i) now=pre; 19 ans+=dfs(pos-1,now,Limit&&(i==End)); 20 } if(pre>=0&&!Limit) dp[pos][pre]=ans; 21 return ans; 22 } 23 signed main() { 24 //freopen("data.in","r",stdin); 25 //freopen("my.out","w",stdout); 26 scanf("%lld",&T); 27 for(int t=1; t<=T; ++t,bit[0]=0) { 28 memset(dp,0xff,sizeof(dp)); 29 scanf("%lld",&n); 30 long long x=n; 31 while(n) bit[++bit[0]]=n%10,n/=10; 32 printf("%lld\n",x-dfs(bit[0],-10,true)); 33 } return 0; 34 }
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