hdu 4734 F(x)(数位dp)
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Problem Description
For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 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 < 109)
For each test case, there are two numbers A and B (0 <= A,B < 109)
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
题意:题目给了个f(x)的定义:F(x) = An * 2n-1 + An-1 * 2n-2 + … + A2 * 2 + A1 * 1,Ai是十进制数位,然后给出a,b求区间[0,b]内满足f(i)==f(a)的i的个数。
思路:最开始的思路是dp[len][sum] 然后每次都memset果断tle 然后觉的开三维 空间又过不了 看聚聚的题解知道可以用总和的思想来减去当前的权值
仔细想想这个状态是与f(a)无关的,一个状态只有在sum=0时才满足,如果我们按常规的思想求f(i)的话,那么最后sum=f(a)才是满足的条件。
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<map> #include<queue> #define ll long long int using namespace std; inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;} int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int dir[2][2]={1,0 ,0,1}; int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1}; const int inf=0x3f3f3f3f; int bits[10]; int dp[10][10007]; //len位 需要凑sum的权值和 int fa; int v[11]={1,2,4,8,16,32,64,128,256,512}; int dfs(int len,int sum,bool ismax){ if(!len) return sum<=fa; if(sum>fa) return 0; if(!ismax&&dp[len][fa-sum]>=0) return dp[len][fa-sum]; int up=ismax?bits[len]:9; int cnt=0; for(int i=0;i<=up;i++){ cnt+=dfs(len-1,sum+i*v[len-1],ismax&&i==up); } if(!ismax) dp[len][fa-sum]=cnt; return cnt; } int solve(int x){ int len=0; while(x){ bits[++len]=x%10; x/=10; } return dfs(len,0,true); } void cal(int a){ int now=1; while(a){ fa+=now*(a%10); a/=10; now*=2; } } int main(){ int t; scanf("%d",&t); int w=0; memset(dp,-1,sizeof(dp)); while(t--){ fa=0; int a,b; scanf("%d%d",&a,&b); cal(a); printf("Case #%d: %d\n",++w,solve(b)); } return 0; }
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