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].
思路：
枚举中心轴的位置，然后针对这个位置，算出结果，结果记得去掉全0的情况。
View Code
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 using std::swap; 6 using std::cin; 7 using std::cout; 8 using std::endl; 9 typedef long long LL; 10 int const N = 25; 11 int const M = 200; 12 LL dp[N][N][M]; 13 int bit[N],ll; 14 LL getsum1(int t,int pre,int center,int limit) 15 { 16 if(!t)return pre==0; 17 if(pre<0)return 0; 18 if(!limit&&dp[t][center][pre]!=-1)return dp[t][center][pre]; 19 int up=(limit?bit[t]:9); 20 LL ans=0; 21 for(int i=0;i<=up;i++) 22 { 23 ans+=getsum1(t-1,pre+(t-center)*i,center,limit&&(i==up)); 24 } 25 if(!limit&&dp[t][center][pre]==-1)dp[t][center][pre]=ans; 26 return ans; 27 } 28 LL getsum2(LL n) 29 { 30 if(n<0)return 0; 31 for(ll=0;n;n/=10)bit[++ll]=n%10; 32 LL ans=0; 33 for(int i=1;i<=ll;i++) 34 ans+=getsum1(ll,0,i,1); 35 return ans-ll+1; 36 } 37 int main() 38 { 39 int T; 40 memset(dp,-1,sizeof(dp)); 41 scanf("%d",&T); 42 LL l,r; 43 while(T--) 44 { 45 scanf("%lld %lld",&l,&r); 46 printf("%lld\n",getsum2(r)-getsum2(l-1)); 47 } 48 return 0; 49 }