HDU-4722 Good Numbers 数位DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4722

　　简单的数位DP，f[i][j][k]表示第 i 位数为 j 时余数为k的个数，然后直接找就可以了。。

  1 //STATUS:C++_AC_31MS_412KB

  2 #include <algorithm>

  3 #include <iostream>

  4 //#include <ext/rope>

  5 #include <fstream>

  6 #include <sstream>

  7 #include <iomanip>

  8 #include <numeric>

  9 #include <cstring>

 10 #include <cassert>

 11 #include <cstdio>

 12 #include <string>

 13 #include <vector>

 14 #include <bitset>

 15 #include <queue>

 16 #include <stack>

 17 #include <cmath>

 18 #include <ctime>

 19 #include <list>

 20 #include <set>

 21 #include <map>

 22 using namespace std;

 23 //#pragma comment(linker,"/STACK:102400000,102400000")

 24 //using namespace __gnu_cxx;

 25 //define

 26 #define pii pair<int,int>

 27 #define mem(a,b) memset(a,b,sizeof(a))

 28 #define lson l,mid,rt<<1

 29 #define rson mid+1,r,rt<<1|1

 30 #define PI acos(-1.0)

 31 //typedef

 32 typedef __int64 LL;

 33 typedef unsigned __int64 ULL;

 34 //const

 35 const int N=25;

 36 const int INF=0x3f3f3f3f;

 37 const int MOD=1000000007,STA=8000010;

 38 const LL LNF=1LL<<60;

 39 const double EPS=1e-8;

 40 const double OO=1e15;

 41 const int dx[4]={-1,0,1,0};

 42 const int dy[4]={0,1,0,-1};

 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 44 //Daily Use ...

 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 55 //End

 56 

 57 LL f[N][N][N];

 58 LL a,b;

 59 int num[N];

 60 int T;

 61 

 62 int getnum(LL n)

 63 {

 64     int i,len=0;

 65     while(n){

 66         num[++len]=n%10;

 67         n/=10;

 68     }

 69     return len;

 70 }

 71 

 72 LL getans(LL n)

 73 {

 74     int i,j,len,sum=0;

 75     LL ret=0;

 76     len=getnum(n);

 77     for(i=len;i>=1;i--){

 78         for(j=0;j<=num[i];j++){

 79             if(j==num[i] && i!=1)continue;

 80             ret+=f[i][j][(10-sum)%10];

 81         }

 82         sum=(sum+num[i])%10;

 83     }

 84     return ret;

 85 }

 86 

 87 int main(){

 88  //   freopen("in.txt","r",stdin);

 89     int i,j,k,p,q,ca=1;

 90     LL ansb,ansa;

 91     mem(f,0);f[0][0][0]=1;

 92     for(i=1;i<=19;i++){

 93         for(j=0;j<=9;j++){

 94             for(k=0;k<=9;k++){

 95                 for(p=0;p<=9;p++){

 96                     f[i][j][k]+=f[i-1][p][(k-j+10)%10];

 97                 }

 98             }

 99         }

100     }

101     scanf("%d",&T);

102     while(T--)

103     {

104         scanf("%I64d%I64d",&a,&b);

105 

106         ansb=(b>0)?getans(b):1;

107         ansa=(a-1>0)?getans(a-1):a;

108 

109         printf("Case #%d: %I64d\n",ca++,ansb-ansa);

110     }

111     return 0;

112 }