hdu 3709 Balanced Number (数位DP)
来源:互联网 发布:淘宝代付可以用信用卡 编辑:程序博客网 时间:2024/05/28 11:30
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].
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
2
0 9
7604 24324
Sample Output
10
897
*题意:找出区间内平衡数的个数,所谓的平衡数
平衡数的概念:数n以数n中的某个位为支点,每个位上的数权值为(数字xi*(posi - 支点的posi)),如果数n里有一个支点使得所有数权值之和为0那么她就是平衡数。比如4139,以3为支点,左边 = 4 * (4 - 2) + 1 * (3 - 2) = 9,右边 = 9 * (1 - 2) = -9,左边加右边为0,所以4139是平衡数。*
*思路:数位DP+记忆化搜索;对于以哪位为支点,我们可以枚举,但需要注意每一次枚举都有0,我们最后得到结果需要减去0的重复数量;
dp[i][j][k] i表示处理到的数位,j是支点,k是力矩和。但是要要把全是0的数排除*
代码:
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;__int64 dp[20][20][2000];int b[20];__int64 dfs(int len,int x,int pos,int flag){ if(len==0) return x==0; if(x<0) return 0; if(!flag&&dp[len][pos][x]!=-1) return dp[len][pos][x]; __int64 ans=0; int end=flag?b[len]:9; for(int i=0;i<=end;i++) { ans+=dfs(len-1,x+(len-pos)*i,pos,flag&&i==end); } if(!flag) dp[len][pos][x]=ans; return ans;}__int64 solve(__int64 n){ int len=0; while(n) { b[++len]=n%10; n=n/10; } __int64 ans=0; for(int i=1;i<=len;i++) { ans+=dfs(len,0,i,1); } return ans-len;}int main(){ int t; scanf("%d",&t); memset(dp,-1,sizeof(dp)); while(t--) { __int64 n,m; scanf("%I64d%I64d",&n,&m); printf("%I64d\n",solve(m)-solve(n-1)); }}
- HDU 3709 Balanced Number(数位DP)
- hdu 3709 Balanced Number (数位dp)
- hdu 3709 Balanced Number(数位dp)
- HDU 3709 Balanced Number(数位DP)
- HDU 3709 Balanced Number (数位DP)
- HDU 3709 Balanced Number (数位dp)
- HDU-3709 Balanced Number (数位DP)
- HDU 3709 Balanced Number(数位dp)
- hdu 3709 Balanced Number (数位DP)
- HDU 3709 Balanced Number (数位DP)
- HDU 3709 Balanced Number(数位dp)
- HDU 3709 Balanced Number(数位DP)
- HDU 3709 Balanced Number(数位dp)
- HDU-3709 Balanced Number (数位dp)
- HDU 3709 Balanced Number(数位DP)
- 数位dp HDU 3709 Balanced Number
- Balanced Number - HDU 3709 数位dp
- [数位dp] hdu 3709 Balanced Number
- Android自定义View年龄范围选择器
- 第四周项目2—— 建设“单链表”算法库
- V尼熊的java之旅——关键字篇
- 十二、oracle 数据库(表)的逻辑备份与恢复
- 2016 JAVA与Android面试题整理
- hdu 3709 Balanced Number (数位DP)
- 第三周项目(2)
- 【Java深入学习系列】之CPU的分支预测(Branch Prediction)模型
- Caffe 初学拾遗(七) Layer Catalogue (Vision Layer)
- LeetCode 389 Find the Difference (异或)
- android ListView分页加载
- 十三、oracle 数据字典和动态性能视图
- Maven 开 发 规 范
- 第4周项目2-建立“单链表”算法库