Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15102 Accepted Submission(s): 5452
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
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.Author
fatboy_cw@WHU
Source
/*
这个是自己的代码反着求的，正如“不要49”，跑了65ms
*/
#include<iostream> #include<stdio.h> #include<vector> #include<algorithm> #include<string.h> #include<cstdio> #define N 22 using namespace std; long long t,n; int g[N];//用来存放数字的位数 long long dp[N][N];//dp[i][j][k]表示当前剩余位数是i上一位数字是j且当前位上是否为4的状态的个数 long long dfs(int len,bool s,bool f)//s表示是不是当前位上是不是4 { if(len<1) return 1; if(!f&&dp[len][s]!=-1) return dp[len][s]; int fmax=f?g[len]:9; long long cur=0; //cout<<"fmax="<<fmax<<endl; for(int i=0;i<=fmax;i++) { if(s&&i==9) continue;//不要有49 cur+=dfs(len-1,i==4,f&&(i==fmax)); //cout<<"cur="<<cur<<endl; } //cout<<cur<<endl; if(!f) dp[len][s]=cur; return cur; } long long solve(long long x) { int len=1; while(x!=0) { g[len++]=x%10; x/=10; } //for(int i=1;i<=len;i++) // cout<<g[i]; //cout<<endl; memset(dp,-1,sizeof dp); return dfs(len-1,false,true); } int main() { //freopen("in.txt","r",stdin); scanf("%lld",&t); while(t--) { scanf("%I64d",&n); printf("%I64d ",n-solve(n)+1); } return 0; }
/* 一开始真的按照“不要49”这么来求的。写博客的时候看看别人博客吸收精华，下面是正着求得，z[i]数组真是神来之笔 本来我也行正着求但是不知道怎么表示找到49之后应该加什么，一个z[i]完美解决了找到49之后应该加多少 */ #include<iostream> #include<stdio.h> #include<vector> #include<algorithm> #include<string.h> #include<cstdio> #define N 30 using namespace std; long long t,n; int g[N];//用来存放数字的位数 long long dp[N][2];//dp[i][j][k]表示当前剩余位数是i上一位数字是j且当前位上是否为4的状态的个数 long long z[N]={1}; long long dfs(int len,bool s,bool f)//s表示是不是当前位上是不是4 { if(len==0) return 0; if(!f&&dp[len][s]>=0) return dp[len][s]; int fmax=f?g[len]:9; long long cur=0; //cout<<"fmax="<<fmax<<endl; for(int i=0;i<=fmax;i++) { if(s&&i==9) { cur+=f?n%z[len-1]+1:z[len-1];//当前位找到了，剩下的不用搜了，直接加上就行了，这里加的是剩下的所有位 } else cur+=dfs(len-1,i==4,f&&g[len]==i); //cout<<"cur="<<cur<<endl; } //cout<<cur<<endl; return f?cur:dp[len][s]=cur; } long long solve(long long x) { int len=0; while(x) { g[++len]=x%10; x/=10; } //for(int i=1;i<=len;i++) // cout<<g[i]; //cout<<endl; g[len+1]=0; return dfs(len,false,true); } int main() { //freopen("in.txt","r",stdin); for (int i=1;i<N;i++) { z[i]=z[i-1]*10; } scanf("%lld",&t); memset(dp,-1,sizeof dp); while(t--) { scanf("%lld",&n); printf("%lld ",solve(n)); } return 0; }