HDU3652--容斥+数位dp
题意：含13子串且整除13的个数
思路：
n-不含13-不整除13+既不也不13（容斥）
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr strcat 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 #include <iomanip> 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 25 //****************** 26 clock_t __START,__END; 27 double __TOTALTIME; 28 void _MS(){__START=clock();} 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 30 //*********************** 31 #define rint register int 32 #define fo(a,b,c) for(rint a=b;a<=c;++a) 33 #define fr(a,b,c) for(rint a=b;a>=c;--a) 34 #define mem(a,b) memset(a,b,sizeof(a)) 35 #define pr printf 36 #define sc scanf 37 #define ls rt<<1 38 #define rs rt<<1|1 39 typedef pair<int,int> PII; 40 typedef vector<int> VI; 41 typedef unsigned long long ull; 42 typedef long long ll; 43 typedef double db; 44 const db E=2.718281828; 45 const db PI=acos(-1.0); 46 const ll INF=(1LL<<60); 47 const int inf=(1<<30); 48 const db ESP=1e-9; 49 const int mod=(int)1e9+7; 50 const int N=(int)1e6+10; 51 52 //开头为i,长度为j,%13为k; 53 int dp[10][15][15]; 54 ll ten[12]; 55 56 void Init(int len) 57 { 58 ten[1]=1; 59 for(int i=2;i<=12;++i)ten[i]=ten[i-1]*10; 60 61 dp[0][0][0]=1; 62 63 for(int i=1;i<=len;++i) 64 { 65 for(int now=0;now<=9;++now) 66 { 67 for(int pre=0;pre<=9;++pre) 68 { 69 for(int premain=0;premain<=12;++premain) 70 { 71 if(now==1&&pre==3)continue; 72 dp[now][i][(ten[i]*now+premain)%13]+=dp[pre][i-1][premain]; 73 } 74 } 75 } 76 } 77 } 78 79 int clac(int x) 80 { 81 int tx=x-1; 82 int num[15]; 83 int ans=0,cnt=0;//既不也不的 84 int ans2=0;//不含13的 85 int ans1=tx-tx/13;//%13!=0的 86 while(x) 87 num[++cnt]=x%10,x/=10; 88 num[cnt+1]=0; 89 ll temp=0; 90 for(int i=cnt;i>=1;--i) 91 { 92 temp+=num[i+1]*ten[i+1]; 93 for(int j=0;j<num[i];++j) 94 { 95 for(int k=0;k<=12;++k) 96 { 97 if((temp+k)%13!=0&&!(j==3&&num[i+1]==1)) 98 ans+=dp[j][i][k]; 99 if(j==3&&num[i+1]==1)continue; 100 ans2+=dp[j][i][k]; 101 } 102 } 103 if(num[i]==3&&num[i+1]==1)break; 104 } 105 ans2--; 106 return tx-ans1-ans2+ans; 107 } 108 109 bool J(int x) 110 { 111 int num[15]; 112 int cnt=0; 113 while(x) 114 num[++cnt]=x%10,x/=10; 115 for(int i=cnt;i>1;--i) 116 if(num[i]==1&&num[i-1]==3)return 1; 117 return 0; 118 } 119 120 int main() 121 { 122 Init(10); 123 int n; 124 while(~sc("%d",&n)) 125 pr("%d\n",clac(n+1)); 126 return 0; 127 } 128 129 /**************************************************************************************/
