HDU 3652 数位DP

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
#include <cassert>
#include <complex>
using namespace std;
typedef long long ll;
typedef long double ld;
const int int_max = 0x07777777;
const int int_min = 0x80000000;
const int inf=0x20202020;
const ll mod=1000000007;
const double eps=1e-9;
const double pi=3.1415926535897932384626;
const int DX[]={1,0,-1,0},DY[]={0,1,0,-1};
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
bool cmp (const void *a , const void *b )
{
return *(int *)a < *(int *)b;
}
int n;
int dp[12][10][2][13];    //dp[i][j][k][l] 表示位数的i，最高位为j，k（0表示不存在13,1表示存在13），l表示被13取模的结果 ，满足上述条件的数的个数。
int a[12];
int solve (int len){
int sum = 0, tempsum=0, flag = 0;
ll t[11];
t[1] = 1;
for(int i = 2; i < 11; i++) t[i] = 10*t[i-1];
for(int i = len; i >= 1; i--){
for(int x = 0; x < a[i]; x++){
int lll = 13-tempsum;
lll %= 13;
lll = lll < 0 ? lll+13 : lll;
sum += dp[i][x][1][lll];
if(flag || (x==3 && a[i+1]==1)){
sum += dp[i][x][0][lll];
}
}
if(a[i]==3&&a[i+1]==1) flag = 1;
tempsum += a[i]*t[i];
}
return sum;
}

int main()
{
memset(dp,0,sizeof(dp));
dp[0][0][0][0] = 1;
for(int i = 1; i < 12; i++){
for(int j = 0; j < 10; j++){
for(int k = 0; k < 2; k++){
for(int l = 0; l < 13; l++){
for(int x = 0; x < 10; x++){
if(k==0){
if(j==1 && x==3) continue;
int lll = l - j*(int)pow(10,i-1);
lll %= 13;
lll = lll < 0 ? lll+13 : lll;
dp[i][j][k][l] += dp[i-1][x][k][lll];
}else{
int lll = l - j*(int)pow(10,i-1);
lll %= 13;
lll = lll < 0 ? lll+13 : lll;
dp[i][j][k][l] += dp[i-1][x][k][lll];
if(j==1 &&x==3) dp[i][j][k][l] += dp[i-1][x][0][lll];
}
}
}
}
}
}

while(scanf("%d", &n)!=EOF){
memset(a,0,sizeof(a));
n++;
int len = 0;
while(n){
a[++len] = n%10;
n /= 10;
}
int ret = solve(len);
printf("%d\n", ret);
}
}