题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3652
题意：找区间内的数，数内含有连续的13并且这个数能被13整除。
dfs(int l, bool one, bool three, int sum, bool flag, bool ok),分别记录位数，上一位1是否出现，当前位3是否出现，这个数%13的值，是否到达边界以及这条dfs链上的结果是否已经有13了。状态设计得有点傻B了。
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define fr first 4 #define sc second 5 #define cl clear 6 #define BUG puts("here!!!") 7 #define W(a) while(a--) 8 #define pb(a) push_back(a) 9 #define Rint(a) scanf("%d", &a) 10 #define Rll(a) scanf("%I64d", &a) 11 #define Rs(a) scanf("%s", a) 12 #define Cin(a) cin >> a 13 #define FRead() freopen("in", "r", stdin) 14 #define FWrite() freopen("out", "w", stdout) 15 #define Rep(i, len) for(int i = 0; i < (len); i++) 16 #define For(i, a, len) for(int i = (a); i < (len); i++) 17 #define Cls(a) memset((a), 0, sizeof(a)) 18 #define Clr(a, x) memset((a), (x), sizeof(a)) 19 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 20 #define lrt rt << 1 21 #define rrt rt << 1 | 1 22 #define pi 3.14159265359 23 #define RT return 24 #define lowbit(x) x & (-x) 25 #define onecnt(x) __builtin_popcount(x) 26 typedef long long LL; 27 typedef long double LD; 28 typedef unsigned long long ULL; 29 typedef pair<int, int> pii; 30 typedef pair<string, int> psi; 31 typedef pair<LL, LL> pll; 32 typedef map<string, int> msi; 33 typedef vector<int> vi; 34 typedef vector<LL> vl; 35 typedef vector<vl> vvl; 36 typedef vector<bool> vb; 37 38 const int maxn = 15; 39 int digit[maxn]; 40 LL dp[maxn][2][2][15][2]; 41 LL n; 42 43 LL dfs(int l, bool one, bool three, int sum, bool flag, bool ok) { 44 if(l == 0) { 45 if(sum != 0) return 0; 46 if(ok) return 1; 47 return 0; 48 } 49 if(!flag && ~dp[l][one][three][sum][ok]) return dp[l][one][three][sum][ok]; 50 LL ret = 0; 51 int pos = flag ? digit[l] : 9; 52 Rep(i, pos+1) { 53 if(one && i == 3) ret += dfs(l-1, one, true, (sum*10+i)%13, flag&&(i==pos), true); 54 else if(i == 1) ret += dfs(l-1, true, false, (sum*10+i)%13, flag&&(i==pos), ok); 55 else ret += dfs(l-1, false, false, (sum*10+i)%13, flag&&(i==pos), ok); 56 } 57 if(!flag) dp[l][one][three][sum][ok] = ret; 58 return ret; 59 } 60 61 LL f(LL x) { 62 int pos = 0; 63 while(x) { 64 digit[++pos] = x % 10; 65 x /= 10; 66 } 67 return dfs(pos, false, false, 0, true, false); 68 } 69 70 signed main() { 71 //FRead(); 72 Clr(dp, -1); 73 while(cin >> n) { 74 cout << f(n) << endl; 75 } 76 RT 0; 77 }
重写了一个，换了一种状态：
dp(l,one,ok,sum)表示长为l的时候，上一位是否是one，并且判断当前是不是已经是13连着出现，和%13的值。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef long long LL; 5 const int maxn = 22; 6 LL dp[maxn][2][2][maxn]; 7 LL n; 8 int digit[maxn]; 9 10 LL dfs(int l, int one, int ok, int sum, bool flag) { 11 if(l == 0) { 12 if(sum != 0) return 0; 13 if(ok) return 1; 14 return 0; 15 } 16 if(!flag && ~dp[l][one][ok][sum]) return dp[l][one][ok][sum]; 17 int pos = flag ? digit[l] : 9; 18 LL ret = 0; 19 for(int i = 0; i <= pos; i++) { 20 if(one && i == 3) ret += dfs(l-1, false, true, (sum*10+i)%13, flag&&(i==pos)); 21 else if(i == 1) ret += dfs(l-1, true, ok, (sum*10+i)%13, flag&&(i==pos)); 22 else ret += dfs(l-1, false, ok, (sum*10+i)%13, flag&&(i==pos)); 23 } 24 if(!flag) dp[l][one][ok][sum] = ret; 25 return ret; 26 } 27 28 LL f(LL x) { 29 int pos = 0; 30 while(x) { 31 digit[++pos] = x % 10; 32 x /= 10; 33 } 34 return dfs(pos, false, false, 0, true); 35 } 36 37 int main() { 38 // freopen("in", "r", stdin); 39 memset(dp, -1, sizeof(dp)); 40 while(~scanf("%lld", &n)) { 41 cout << f(n) << endl; 42 } 43 return 0; 44 }