【HDU4734】F(x) 数位DP

题目描述

　　For a decimal number x with n digits (A[n]A[n-1]A[n-2] … A[2]A[1]), we define its weight as F(x) = A[n] * 2^(n-1) + A[n-1] * 2^(n-2) + … + A[2] * 2 + A[1] * 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).

题目大意

　　对于一个数字x，n为x的位数，ai为从右往左第i位的数字。f(x)=∑ni=12i−1∗a[i]，给定数字A，B，求在0-B中这B+1个数中，f(x) < f(A) 的x个数。

数据范围

(0 <= A,B < 10^9)

样例输入

3
0 100
1 10
5 100

样例输出

Case #1: 1
Case #2: 2
Case #3: 13

解题思路

数位DP
太水了就不写具体了qwq

代码

#include 
using namespace std;
inline int Getint(){int x=0,f=1;char ch=getchar();while('0'>ch||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while('0'<=ch&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
int A,B,Max,lim[15];
int dp[3005][15];
void f(int x){
    int Len=0,ret=0;
    while(x)ret+=(x%10)*(1<x/=10;
    Max=ret;
}
int Ask(int pos,int Sum,bool flag){
    if(!pos)return Sum<=Max;
    if(!flag&&~dp[Sum][pos])return dp[Sum][pos];
    int bound=flag?lim[pos]:9,ret=0;
    for(int i=0;i<=bound;i++)ret+=Ask(pos-1,Sum+(1<<pos-1)*i,flag&&i==bound);
    return !flag?dp[Sum][pos]=ret:ret;
}
int Ask(int x){
    int Len=0;
    while(x){lim[++Len]=x%10;x/=10;}
    return Ask(Len,0,1);
}
int main(){
    int Case=Getint();
    for(int i=1;i<=Case;i++){
        memset(dp,-1,sizeof(dp));
        f(Getint());
        cout<<"Case #"<": "<"\n";
    }
    return 0;
}

转载于:https://www.cnblogs.com/Cedric341561/p/6810989.html