HDU 4734 F(x)(数位DP)
2015-03-30 21:23
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Description
For a decimal number x with n digits (A nA n-1A n-2 ... A 2A 1), we define its weight as F(x) = A n * 2 n-1 + A n-1 * 2 n-2 + ... +
A 2 * 2 + A1 * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).
Input
The first line has a number T (T <= 10000) , indicating the number of test cases.
For each test case, there are two numbers A and B (0 <= A,B < 10 9)
Output
For every case,you should output "Case #t: " at first, without quotes. The t is the case number starting from 1. Then output the answer.
Sample Input
Sample Output
简单的数位DP:先将A表示成F(x)然后统计有多少大于x就行了。
For a decimal number x with n digits (A nA n-1A n-2 ... A 2A 1), we define its weight as F(x) = A n * 2 n-1 + A n-1 * 2 n-2 + ... +
A 2 * 2 + A1 * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).
Input
The first line has a number T (T <= 10000) , indicating the number of test cases.
For each test case, there are two numbers A and B (0 <= A,B < 10 9)
Output
For every case,you should output "Case #t: " at first, without quotes. The t is the case number starting from 1. Then output the answer.
Sample Input
3 0 100 1 10 5 100
Sample Output
Case #1: 1 Case #2: 2 Case #3: 13
简单的数位DP:先将A表示成F(x)然后统计有多少大于x就行了。
#include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<string> #include<iostream> #include<queue> #include<cmath> #include<map> #include<stack> #include<bitset> using namespace std; #define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define REP( i , n ) for ( int i = 0 ; i < n ; ++ i ) #define CLEAR( a , x ) memset ( a , x , sizeof a ) typedef long long LL; typedef pair<int,int>pil; const int INF = 0x3f3f3f3f; int t,temp; LL l,r; int num[20]; LL dp[20][20000]; LL dfs(int pos,int sum,int flag) { if(pos==0) return sum>=0; if(sum<0) return 0; if(!flag&&dp[pos][sum]!=-1) return dp[pos][sum]; LL ans=0; int ed=flag?num[pos]:9; for(int i=0;i<=ed;i++) { int s=sum-i*(1<<(pos-1)); ans+=dfs(pos-1,s,flag&&i==ed); } if(!flag) dp[pos][sum]=ans; return ans; } LL solve(LL x) { int pos=0; while(x) { num[++pos]=x%10; x/=10; } return dfs(pos,temp,1); } LL work(LL x) { LL ans=0; int l=0; while(x) { ans+=(x%10)*(1<<l); l++;x/=10; } return ans; } int main() { int cas=1;CLEAR(dp,-1); scanf("%d",&t); while(t--) { scanf("%lld%lld",&l,&r); temp=work(l); LL ans=solve(r); printf("Case #%d: %lld\n",cas++,ans); } return 0; }
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