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 }
