hdu3652 数位DP

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题意求小于n是13的倍数且含有13的数的个数。

思路 f(i,j,k,l)代表，i位数中第一位是j的，是否有包含13(k == 1 or 0)，模13余数是l的数有几个。预处理时，要注意枚举i-1位数不同的余数。详见代码~

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxlen = 11;int f[maxlen][10][2][13];int pow10(int x){int ret = 1;for(int i=0;i<x;i++)ret *= 10;return ret;}void yu(){f[0][0][0][0] = 1;for(int i=1;i<maxlen;i++){for(int j=0;j<=9;j++){if(i==maxlen-1 && j>=2)break;for(int k=0;k<2;k++){for(int l=0;l<13;l++){int& now = f[i][j][k][l];for(int x=0;x<=9;x++){int lpre = (l-(j*pow10(i-1))%13+13)%13;if(k==0){if(x==3 && j==1){continue;}now += f[i-1][x][k][lpre];}else{if(x==3&&j==1){now += f[i-1][x][0][lpre];}now += f[i-1][x][k][lpre];}}}}}}}int digit[maxlen];int chaifen(int n){memset(digit,0,sizeof(digit));int i = 0;while(n>0){i++;digit[i] = n%10;n/=10;}return i;}//(0,n)int go(int n){int len = chaifen(n);bool flag = 0;int ans = 0;int now = 0;for(int i=len;i>=1;i--){now += digit[i+1]*pow10(i); for(int j=0;j<=digit[i]-1;j++){if(flag){ans += f[i][j][0][(13-now%13)%13] + f[i][j][1][(13-now%13)%13];}else{ans += f[i][j][1][(13-now%13)%13];if(digit[i+1] == 1 && j==3)ans += f[i][j][0][(13-now%13)%13];}}if(digit[i+1] == 1&& digit[i] == 3)flag = 1;}return ans;}int main(){//freopen("out.txt","w",stdout);yu();int n;while(scanf("%d",&n) == 1){printf("%d\n",go(n+1));}return 0;}

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