HDU 3652 数位DP(递归)
2013-04-15 14:47
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递归
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int a[11];
int n;
/*
* 0 包含13
* 1 不包含13,但以3开头
* 2 不包含13,不以3开头
*/
int dp[11][3][15];
int dfs(int len, int pre, int mod, bool lim, bool has) {
if(!len) return has && !mod;
if(!lim) {
if(!has && pre != 1 && ~dp[len][0][mod]) return dp[len][0][mod];
if(!has && pre == 1 && ~dp[len][1][mod]) return dp[len][1][mod];
if(has && ~dp[len][2][mod]) return dp[len][2][mod];
}
int m = lim ? a[len-1] : 9;
int ret = 0;
for(int i = 0; i <= m; i++) {
ret += dfs(len-1, i, (mod*10+i)%13, lim&&(i==m), has||(pre == 1 &&i==3));
}
if(!lim){
if(has) dp[len][2][mod] = ret;
if(!has && pre != 1) dp[len][0][mod] = ret;
if(!has && pre == 1) dp[len][1][mod] = ret;
}
return ret;
}
int gao(int n) {
int num = 0;
while(n) {
a[num++] = n % 10;
n /= 10;
}
memset(dp, -1, sizeof(dp));
return dfs(num, 0, 0, 1, 0);
}
int main() {
while(cin >> n) cout << gao(n) << endl;
return 0;
}
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int a[11];
int n;
/*
* 0 包含13
* 1 不包含13,但以3开头
* 2 不包含13,不以3开头
*/
int dp[11][3][15];
int dfs(int len, int pre, int mod, bool lim, bool has) {
if(!len) return has && !mod;
if(!lim) {
if(!has && pre != 1 && ~dp[len][0][mod]) return dp[len][0][mod];
if(!has && pre == 1 && ~dp[len][1][mod]) return dp[len][1][mod];
if(has && ~dp[len][2][mod]) return dp[len][2][mod];
}
int m = lim ? a[len-1] : 9;
int ret = 0;
for(int i = 0; i <= m; i++) {
ret += dfs(len-1, i, (mod*10+i)%13, lim&&(i==m), has||(pre == 1 &&i==3));
}
if(!lim){
if(has) dp[len][2][mod] = ret;
if(!has && pre != 1) dp[len][0][mod] = ret;
if(!has && pre == 1) dp[len][1][mod] = ret;
}
return ret;
}
int gao(int n) {
int num = 0;
while(n) {
a[num++] = n % 10;
n /= 10;
}
memset(dp, -1, sizeof(dp));
return dfs(num, 0, 0, 1, 0);
}
int main() {
while(cin >> n) cout << gao(n) << endl;
return 0;
}
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