hdu3709balanced number Digit DP

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Balanced Number Time limit:10000/5000 MS (java/others) Memory limit:65535/65535 K (java/others) Total submission (s): 3549 Accepted Submission (s): 1630 Problem Description A balanced number is a non-negative integer the can be balanced if A pivot was placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot was placed at some digit of the number, the distance from a digit to the pivot is the offset between it and th e pivot. Then the torques of left part and right part can calculated. It's balanced if they is the same. A balanced number must is balanced with the pivot at some of its digits. For example, 4139 are a balanced number with pivot fixed at 3. The torqueses is 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 was the total number of cases T (0 < t≤30). For each case, there is integers separated by a space in a line, X and Y. (0≤x≤y≤10 18). Output for each case, print the number of balanced numbers in the range [x, y] with a line. Sample Input 2 0 9) 7604 24324 Sample Output 10 897 Author GAO, Yuan Source of Asia Chengdu regional Contest

Hold a noon only a dropped the question = = also is a digital DP in the more classic

Say test instructions: Given a known number string, requiring a number to be centered on both sides of the moment and the number of equal, we consider three results related to the amount: the length of the current enumeration, the position of the axis, the moment and, respectively, as a DP of three-dimensional

Flag to mark whether the maximum bit is used, notice sum<0 exit; pos=0, if Sum=0 then the number of methods equal to +1, because the requirement is ==0, that is, in the process of recursion, POS center of the relative position is uncertain, POS move to the right is still pos-center so is added negative value = = to the last flag=0, we can assume that the current situation is completed before the DP can be assigned

Understand this problem well to make it to the next poj3252--

/************ hdu3709 2016.3.11 31MS 22860K 1075 B C + + ************/#include <iostream> #include <cstdio> #inc
Lude<cstring> using namespace std;
Long Long dp[30][30][3000];
int num[30];
    Long Long dfs (int pos,int center,int Sum,bool flag) {if (pos==0) return sum==0;
    if (sum<0) return 0;
    if (!flag&&dp[pos][center][sum]!=-1) return dp[pos][center][sum];
    Long Long ans=0;
    int maxn=flag?num[pos]:9;
    for (int i=0;i<=maxn;i++) {Ans+=dfs (pos-1,center,sum+i* (pos-center), I==maxn&&flag);
    }//pos-center if (!flag) Dp[pos][center][sum]=ans;
return ans;
    } A long long cal (Long long x) {long long ans=0;
    int pos=0;
        while (x) {num[++pos]=x%10;
    x/=10;
    } for (int i=1;i<=pos;i++) Ans+=dfs (pos,i,0,1);
return ans-pos+2;
    } int main () {//freopen ("Cin.txt", "R", stdin);
    int t;
    Long Long n,m;
    Memset (Dp,-1,sizeof (DP));
    scanf ("%d", &t);
      while (t--) {  scanf ("%i64d%i64d", &n,&m);
    printf ("%i64d\n", Cal (M)-cal (n-1));
} return 0; }