2016多校训练Contest5: 1007 K-wolf Number hdu5787

Problem Description

Alice thinks an integer x is a K-wolf number, if every K adjacent digits in decimal representation of x is pairwised different.
Given (L,R,K), please count how many K-wolf numbers in range of [L,R].

 

Input

The input contains multiple test cases. There are about 10 test cases.

Each test case contains three integers L, R and K.

1≤L≤R≤1e18
2≤K≤5

 

Output

For each test case output a line contains an integer.

 

Sample Input

1 1 220 100 5

 

Sample Output

172

直接数位dp

f[i][d1]……[dk]表示前i位前面各位分别为d1……dk的方案数

直接枚举位数转移即可

注意处理前导0和001023这种情况

可以考虑把前导0赋值成10

因为不想传递数组各种判断，就k=2 3 4 5分别写了一份程序

#include<map>#include<cmath>#include<queue>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int a[21];long long f2[21][11][11];int dd[11];inline long long dfs2(int len,int d1,int d2,bool lim,bool lim2){int i,j;dd[1]=d1;dd[2]=d2;bool flag=true;for(i=1;i<=2;i++){if(dd[i]==10)continue;for(j=i+1;j<=2;j++){if(dd[i]==dd[j]){flag=false;break;}}if(!flag) break;}    if(!flag)        return 0;    else if(lim2&&len==0)    return 0;    else if(len==0)    return 1;    if(!lim&&f2[len][d1][d2]!=0)        return f2[len][d1][d2];    int limit;    if(lim)        limit=a[len];    else        limit=9;    long long ans=0;    for(i=0;i<=limit;i++)    {    int dx=i;    if(dx==0&&lim2)    dx=10;        ans+=dfs2(len-1,d2,dx,lim&&i==a[len],lim2&&i==0);    }    if(!lim)        f2[len][d1][d2]=ans;    return ans;}inline long long cale2(long long x){    int len=0;    while(x!=0)    {        len++;        a[len]=x%(long long)10;        x=x/(long long)10;    }    int i,j;    long long ans=0;    ans+=dfs2(len,10,10,true,true);    return ans;}//--------------------------3-----------------------------------long long f3[21][11][11][11];inline long long dfs3(int len,int d1,int d2,int d3,bool lim,bool lim2){int i,j;dd[1]=d1;dd[2]=d2;dd[3]=d3;bool flag=true;for(i=1;i<=3;i++){if(dd[i]==10)continue;for(j=i+1;j<=3;j++){if(dd[i]==dd[j]){flag=false;break;}}if(!flag) break;}    if(!flag)        return 0;    else if(lim2&&len==0)    return 0;    else if(len==0)    return 1;    if(!lim&&f3[len][d1][d2][d3]!=0)        return f3[len][d1][d2][d3];    int limit;    if(lim)        limit=a[len];    else        limit=9;    long long ans=0;    for(i=0;i<=limit;i++)    {    int dx=i;    if(dx==0&&lim2)    dx=10;        ans+=dfs3(len-1,d2,d3,dx,lim&&i==a[len],lim2&&i==0);    }    if(!lim)        f3[len][d1][d2][d3]=ans;    return ans;}inline long long cale3(long long x){    int len=0;    while(x!=0)    {        len++;        a[len]=x%(long long)10;        x=x/(long long)10;    }    int i,j;    long long ans=0;    ans+=dfs3(len,10,10,10,true,true);    return ans;}//--------------------------4-----------------------------------long long f4[21][11][11][11][11];inline long long dfs4(int len,int d1,int d2,int d3,int d4,bool lim,bool lim2){int i,j;dd[1]=d1;dd[2]=d2;dd[3]=d3;dd[4]=d4;bool flag=true;for(i=1;i<=4;i++){if(dd[i]==10)continue;for(j=i+1;j<=4;j++){if(dd[i]==dd[j]){flag=false;break;}}if(!flag) break;}    if(!flag)        return 0;    else if(lim2&&len==0)    return 0;    else if(len==0)    return 1;    if(!lim&&f4[len][d1][d2][d3][d4]!=0)        return f4[len][d1][d2][d3][d4];    int limit;    if(lim)        limit=a[len];    else        limit=9;    long long ans=0;    for(i=0;i<=limit;i++)    {    int dx=i;    if(dx==0&&lim2)    dx=10;        ans+=dfs4(len-1,d2,d3,d4,dx,lim&&i==a[len],lim2&&i==0);    }    if(!lim)        f4[len][d1][d2][d3][d4]=ans;    return ans;}inline long long cale4(long long x){    int len=0;    while(x!=0)    {        len++;        a[len]=x%(long long)10;        x=x/(long long)10;    }    int i,j;    long long ans=0;    ans+=dfs4(len,10,10,10,10,true,true);    return ans;}//--------------------------5-----------------------------------long long f5[21][11][11][11][11][11];inline long long dfs5(int len,int d1,int d2,int d3,int d4,int d5,bool lim,bool lim2){int i,j;dd[1]=d1;dd[2]=d2;dd[3]=d3;dd[4]=d4;dd[5]=d5;bool flag=true;for(i=1;i<=5;i++){if(dd[i]==10)continue;for(j=i+1;j<=5;j++){if(dd[i]==dd[j]){flag=false;break;}}if(!flag) break;}    if(!flag)        return 0;    else if(lim2&&len==0)    return 0;    else if(len==0)    return 1;    if(!lim&&f5[len][d1][d2][d3][d4][d5]!=0)        return f5[len][d1][d2][d3][d4][d5];    int limit;    if(lim)        limit=a[len];    else        limit=9;    long long ans=0;    for(i=0;i<=limit;i++)    {    int dx=i;    if(dx==0&&lim2)    dx=10;        ans+=dfs5(len-1,d2,d3,d4,d5,dx,lim&&i==a[len],lim2&&i==0);    }    if(!lim)        f5[len][d1][d2][d3][d4][d5]=ans;    return ans;}inline long long cale5(long long x){    int len=0;    while(x!=0)    {        len++;        a[len]=x%(long long)10;        x=x/(long long)10;    }    int i,j;    long long ans=0;    ans+=dfs5(len,10,10,10,10,10,true,true);    return ans;}int main(){long long l,r;int k;while(scanf("%I64d%I64d%d",&l,&r,&k)!=EOF){if(k==2)printf("%I64d\n",cale2(r)-cale2(l-1));else if(k==3)printf("%I64d\n",cale3(r)-cale3(l-1));else if(k==4)printf("%I64d\n",cale4(r)-cale4(l-1));else if(k==5)printf("%I64d\n",cale5(r)-cale5(l-1));}return 0;}