hdoj-【3555 Bomb】

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 16141    Accepted Submission(s): 5897

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

3150500

 

Sample Output

0115
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.
 

 

这是数位dp的第二道题目，不用问又是看的网上的代码，这里主要解决的问题是
当当前位已经满足是49该怎么计算，看z数组，就是剩下的位不用计算直接相加就行了
 temp+=limit?n%f[len-1]+1:f[len-1];估计这个地方也有不懂得同学吧，
 比如49845这个数满足条件这时候不能加1000；因为这样就大于49845，我们应该加上845

就是对1000取余 

One

<span style="font-size:18px;">#include<cstdio>#include<cstring>#define LL __int64LL a[100],f[30];LL n;LL dp[100][2]; LL dfs(LL len,bool sta,bool limit){if(len==0)return 0;if(!limit&&dp[len][sta]!=-1)return dp[len][sta];LL up,i,temp=0;up=limit?a[len]:9;for(i=0;i<=up;++i){if(sta&&i==9)temp+=limit?n%f[len-1]+1:f[len-1];elsetemp+=dfs(len-1,i==4,limit&&up==i); } if(!limit)dp[len][sta]=temp;return temp; } LL solver(LL n){LL len=1;while(n){a[len++]=n%10;n/=10; } return dfs(len-1,0,1); } int main(){LL t,i;f[0]=1; for(i=1;i<30;++i) f[i]=f[i-1]*10;scanf("%I64d",&t);while(t--){memset(dp,-1,sizeof(dp)); scanf("%I64d",&n);printf("%I64d\n",solver(n)); } } </span>

不要49 从反面考虑

Two

<span style="font-size:18px;">#include<cstdio>#include<cstring>#define LL long longLL dp[30][2],a[100]; LL dfs(LL len,bool sta,bool limit){if(len==0)return 1;if(!limit&&dp[len][sta]!=-1)return dp[len][sta];LL up,temp=0,i;up=limit?a[len]:9;for(i=0;i<=up;++i){if(sta&&i==9)continue;temp+=dfs(len-1,i==4,limit&&i==up); } if(!limit) dp[len][sta]=temp; return temp; } LL solver(LL n){int len=1;while(n){a[len++]=n%10;n/=10; } return dfs(len-1,0,1);} int main(){LL t;scanf("%lld",&t);while(t--){memset(dp,-1,sizeof(dp));LL n;scanf("%lld",&n);printf("%lld\n",n-solver(n)+1);//为什么加一，因为0算符合条件的，那就多减1 }return 0; } </span>