Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <bitset> #include <map> #include <queue> #include <stack> #include <vector> #define rep(i,m,n) for(i=m;i<=n;i++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define sys system("pause") const int maxn=1e5+10; const int N=5e4+10; const int M=N*10*10; using namespace std; inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} inline void umax(ll &p,ll q){if(p<q)p=q;} inline void umin(ll &p,ll q){if(p>q)p=q;} inline ll read() { ll x=0;int f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } int n,m,k,t,num[20],pos; ll dp[20][20][2000],p,q; ll dfs(int pos,int x,int y,int z) { if(pos<0)return y==0; if(y<0)return 0; if(z&&dp[pos][x][y]!=-1)return dp[pos][x][y]; int now=z?9:num[pos],i; ll ret=0; rep(i,0,now) { ret+=dfs(pos-1,x,y+i*(pos-x),z||i<num[pos]); } return z?dp[pos][x][y]=ret:ret; } ll gao(ll p) { if(p<0)return 0; pos=0; while(p)num[pos++]=p%10,p/=10; ll ret=0; int i; rep(i,0,pos-1)ret+=dfs(pos-1,i,0,0); return ret-pos+1; } int main() { int i,j; memset(dp,-1,sizeof(dp)); scanf("%d",&t); while(t--) { scanf("%lld%lld",&p,&q); printf("%lld\n",gao(q)-gao(p-1)); } return 0; }
原文:http://www.cnblogs.com/dyzll/p/6390817.html