F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5775 Accepted Submission(s): 2184
Problem 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 + 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).
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
Source 2013 ACM/ICPC Asia Regional Chengdu Online 想法：數位dp 題目給了個f(x)的定義：F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 1，Ai是十進制數位，然後給出a,b求區間[0,b]内滿足f(i)==f(a)的i的個數。 正常想：這個f(x)計算就和數位計算是一樣的，就是加了權值，是以dp[pos][sum]，這狀态是基本的。a是題目給定的，f(a)是變化的不過f(a)最大好像是4600的樣子。如果要memset優化就要加一維存f(a)的不同取值，那就是dp[10][4600][4600]，這顯然不合法。 這個時候就要用減法了，dp[pos][sum]，sum不是存目前枚舉的數的字首和(權重的)，而是枚舉到目前pos位，後面還需要湊sum的權值和的個數， 也就是說初始的是時候sum是f(a),枚舉一位就減去這一位在計算f(i)的權值，顯然sum=0時就是滿足的，後面的位數湊足sum位就可以了。 仔細想想這個狀态是與f(a)無關的(新手似乎很難了解)，一個狀态隻有在sum=0時才滿足，如果我們按正常的思想求f(i)的話，那麼最後sum=f(a)才是滿足的條件。 代碼： #include<stdio.h>
#include<string.h>
int xx;
int dp[12][10010];
int a[12];
int f(int n)
{
if(n==0)
return 0;
int ans=f(n/10);
return ans*2+(n%10);
}
int dfs(int pos,int sum,int limit)
{
if(pos<0)
return sum<=xx;
if(sum>xx)
return 0;
if(!limit && dp[pos][xx-sum] != -1)
return dp[pos][xx-sum];
int up = limit ? a[pos] : 9;
int tmp = 0;
for(int i = 0; i <= up; i++)
{
tmp += dfs(pos-1,sum+i*(1<<pos),limit&&(i==a[pos]));
}
if(!limit)
dp[pos][xx-sum] = tmp;
return tmp;
}
int solver(int m)
{
int pos=0;
while(m)
{
a[pos++]=m%10;
m/=10;
}
dfs(pos-1,0,1);
}
int main()
{
int T;
scanf("%d",&T);
int ws=1;
memset(dp,-1,sizeof(dp));
while(T--)
{
int n,m;
scanf("%d%d",&n,&m);
xx=f(n);
printf("Case #%d: %d\n",ws++,solver(m));
}
return 0;
}