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杭电 3555 Bomb
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
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.
第一个数位动规!
!!!
看代码理解的。差点儿相同懂那么点点点了。
AC代码例如以下:
传统模板!
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///递推 46MS 328K #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<cstdio> using namespace std; int main() { __int64 dp[20][3];//0表示该位数中不含49的情况数。1表示不含49但头为9的情况数。2表示含49的情况数 int i,j; memset(dp,0,sizeof dp); dp[0][0]=1; for(i=1;i<=20;i++) { dp[i][0]=dp[i-1][0]*10-dp[i-1][1]; dp[i][1]=dp[i-1][0]; dp[i][2]=dp[i-1][2]*10+dp[i-1][1]; //cout<<dp[i][0]<<" "<<dp[i][1]<<" "<<dp[i][2]<<endl; } int t; cin>>t; __int64 num[30]; __int64 n; while(t--) { memset(num,0,sizeof num); cin>>n; n++;//这一步非常重要 int tt=1; while(n) { num[tt++]=n%10; n/=10; } __int64 ans=0; int flag=0;int last=0; for(i=tt-1;i>=1;i--) { ans+=num[i]*dp[i-1][2]; if(flag==1) ans+=dp[i-1][0]*num[i]; if(flag==0&&num[i]>4) ans+=dp[i-1][1]; if(num[i+1]==4&&num[i]==9) flag=1; } cout<<ans<<endl; } return 0; }
///数位之记忆化搜索 78MS 276K #include<cstdio> #include<iostream> #include<cstring> #define mod 1000000009 #define ll __int64 #define M 100005 using namespace std; ll dp[30][3]; ll num[30]; ll dfs(int pos,int statu,int limit) { int i; if(!pos) return statu==2; if(!limit&&dp[pos][statu]!=0) return dp[pos][statu]; int end=limit?num[pos]:9; ll sum=0; for(i=0;i<=end;i++) { int flag=statu; if(flag==0&&i==4) flag=1; if(flag==1&&i==9) flag=2; if(flag==1&&i!=4&&i!=9) flag=0; sum+=dfs(pos-1,flag,limit&&i==end); } return limit?
sum:(dp[pos][statu]=sum); } ll _49(ll n) { int pos=1; memset(dp,0,sizeof dp); while (n>0) { num[pos++]=n%10; n/=10; } //cout<<pos<<"~~~~~~~~~~~~~"<<endl; return dfs(pos-1,0,1); } int main() { int i,j; int t; scanf("%d",&t); while(t--) { ll n; scanf("%I64d",&n); printf("%I64d\n",_49(n)); } return 0; }
杭电 3555 Bomb