HDU 3709 Balanced Number (数位DP)
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Balanced Number
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 6038 Accepted Submission(s): 2887
Problem Description
A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the number, the distance from a digit to the pivot is the offset between it and the pivot. Then the torques of left part and right part can be calculated. It is balanced if they are the same. A balanced number must be balanced with the pivot at some of its digits. For example, 4139 is a balanced number with pivot fixed at 3. The torqueses are 4*2 + 1*1 = 9 and 9*1 = 9, for left part and right part, respectively. It's your job
to calculate the number of balanced numbers in a given range [x, y].
to calculate the number of balanced numbers in a given range [x, y].
Input
The input contains multiple test cases. The first line is the total number of cases T (0 < T ≤ 30). For each case, there are two integers separated by a space in a line, x and y. (0 ≤ x ≤ y ≤ 1018).
Output
For each case, print the number of balanced numbers in the range [x, y] in a line.
Sample Input
20 97604 24324
Sample Output
10897
Author
GAO, Yuan
Source
2010 Asia Chengdu Regional Contest
Recommend
zhengfeng
枚举支点
dp[i][j][k]表示i+1~len位数字已确定,搜索第i位数字,支点为j,力矩为k的可能性
#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>#define maxn 2520typedef long long ll;using namespace std;ll x,y,dp[20][20][3010];int digit[20];ll dfs(int p,int cent,int bal,int limit){ll ans=0;int num,i;if(p==0)return bal==0;if(bal<0)return 0;if(!limit&&dp[p][cent][bal]!=-1)return dp[p][cent][bal];num=limit?digit[p]:9;for(i=0;i<=num;i++){ans+=dfs(p-1,cent,bal+i*(p-cent),limit&&i==num);;}if(!limit)dp[p][cent][bal]=ans;return ans;}ll work(ll n){int len=0;while(n){digit[++len]=n%10;n/=10;}int i;long long ans=0;for(i=1;i<=len;i++){ans+=dfs(len,i,0,1);}return ans-len+1;}int main (){int t;scanf("%d",&t);memset(dp,-1,sizeof(dp));while(t--){scanf("%lld %lld",&x,&y);printf("%lld\n",work(y)-work(x-1));}return 0;}
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