hdu3709

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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 42 + 11 = 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

    题解

  • 模板题
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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,l,n) for(register int i=l;i<n;++i)
#define ll long long
using namespace std;
const int N=22;
ll dp[N][N][2000];//第i位，枢轴为j的时候和为k时共有多少种情况满足符合要求 
int a[N];

ll dfs(int len,int pos,int limit,int sta){
    if(len==-1)return sta?0:1;
    if(sta<0)return 0;
    if(!limit&&dp[len][pos][sta]!=-1)return dp[len][pos][sta];
    int up=limit?a[len]:9;
    ll tmp=0;
    rep(i,0,up+1){
        tmp+=dfs(len-1,pos,limit&&i==a[len],sta+i*(len-pos));
    }
    if(!limit)dp[len][pos][sta]=tmp;
    return tmp;
}


ll solve(ll x){
    int cnt=0;
    while(x){
        a[cnt++]=x%10;
        x/=10;
    }
    ll ans=0;
    rep(i,0,cnt){
        ans+=dfs(cnt-1,i,1,0);
    }
    return ans-cnt+1;
}

int main(){
    ll l,r;
    int T;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    while(T--){
        scanf("%lld%lld",&l,&r);
        printf("%lld\n",solve(r)-solve(l-1));
    }
    return 0;
}