Balanced Number
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 5276 Accepted Submission(s): 2523
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
2
0 9
7604 24324
Sample Output
10
897
Author
GAO, Yuan
Source
题意：给你一个数number，你可以确定某位，然后如果分成的左右两部分的 sigma（d[i] * | i - fixloc |）相等，则它是 Balanced Number。统计区间 [a,b] 中Balance Number 的个数。
题解：枚举分界点 套路数位dp 注意0 在 数位dp 的 dfs 中，对于 0 的看待，是 len 个0 放一起
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <algorithm> 6 #include <stack> 7 #include <queue> 8 #include <cmath> 9 #include <map> 10 #define ll __int64 11 #define dazhi 2147483647 12 #define bug() printf("!!!!!!!") 13 #define M 100005 14 using namespace std; 15 ll a,b; 16 ll f[30][30][2005]; 17 int bit[30]; 18 ll dp(int pos,int x,int st,int flag) 19 { 20 if(pos==0) return st==0; 21 if(st<0) return 0; 22 if(flag&&f[pos][x][st]!=-1) return f[pos][x][st]; 23 int d=flag?9:bit[pos]; 24 ll re=0; 25 for(int i=0;i<=d;i++) 26 { 27 re+=dp(pos-1,x,st+i*(pos-x),flag||i!=d); 28 } 29 if(flag) f[pos][x][st]=re; 30 return re; 31 } 32 ll fun(ll n) 33 { 34 if(n<0) return 0; 35 int len=0; 36 while(n>0) 37 { 38 bit[++len]=n%10; 39 n/=10; 40 } 41 ll ans=0; 42 for(int i=1;i<=len;i++) 43 { 44 ans+=dp(len,i,0,0); 45 } 46 return ans-len+1; 47 } 48 int main() 49 { 50 int t; 51 scanf("%d",&t); 52 memset(f,-1,sizeof(f)); 53 while(t--) 54 { 55 scanf("%I64d %I64d",&a,&b); 56 printf("%I64d ",fun(b)-fun(a-1)); 57 } 58 return 0; 59 }