数位DP
学了一下怎么写递归，发现确实比较简单；
dp[pos][][]对应dfs()中的参数的状态，记忆化当前状态的值，不用考虑这个状态表示什么意思；
然后就是设计好dfs()中的参数；
hdu 3555 http://acm.hdu.edu.cn/showproblem.php?pid=3555
题意：统计1~n之间含有49的数字的个数；
需要记录当前位置，前一位置放了那个数字，当前是否已经包含49，是否有上界；
dfs(pos,pre,istrue,limit);
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<iostream> 6 using namespace std; 7 const int N = 20; 8 typedef long long LL; 9 int dig[N]; 10 LL dp[N][10][2]; 11 12 LL dfs(int pos,int pre,int istrue,int limit) { 13 if (pos < 0) return istrue; 14 if (!limit && dp[pos][pre][istrue] != -1) 15 return dp[pos][pre][istrue]; 16 int last = limit ? dig[pos] : 9; 17 LL ret = 0; 18 for (int i = 0; i <= last; i++) { 19 ret += dfs(pos-1,i,istrue || (pre == 4 && i == 9),limit && (i == last)); 20 } 21 if (!limit) { 22 dp[pos][pre][istrue] = ret; 23 } 24 return ret; 25 } 26 LL solve(LL n) { 27 int len = 0; 28 while (n) { 29 dig[len++] = n % 10; 30 n /= 10; 31 } 32 return dfs(len-1,0,0,1); 33 } 34 int main(){ 35 memset(dp,-1,sizeof(dp)); 36 int T; scanf("%d",&T); 37 while (T--) { 38 LL n; 39 cin>>n; 40 cout<<solve(n)<<endl; 41 } 42 return 0; 43 }
usetc 1307 http://acm.uestc.edu.cn/problem.php?pid=1307
题意：相邻两个数之差大于2；
dfs(pos,pre,limit,fg); fg表示前面是否全为0
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstdlib> 6 #include<cstring> 7 #include<iostream> 8 using namespace std; 9 typedef long long LL; 10 const int N = 20; 11 LL dp[N][10][2]; 12 LL a,b; 13 int dig[N]; 14 LL dfs(int pos,int pre,int limit,int fg) { 15 if (pos < 0) return 1; 16 if (!limit && dp[pos][pre][fg] != -1) 17 return dp[pos][pre][fg]; 18 int last = limit ? dig[pos] : 9; 19 LL ret = 0; 20 for (int i = 0; i <= last; i++) { 21 if (fg == 0 || abs(i - pre) >= 2) 22 ret += dfs(pos-1,i,limit && (i == last),fg || i); 23 } 24 if (!limit) { 25 dp[pos][pre][fg] = ret; 26 } 27 return ret; 28 } 29 LL solve(LL n) { 30 int len = 0; 31 if (n < 0) return 0; 32 while (n) { 33 dig[len++] = n % 10; 34 n /= 10; 35 } 36 return dfs(len-1,0,1,0); 37 } 38 int main(){ 39 memset(dp,-1,sizeof(dp)); 40 // cout<<solve(15)<<endl; 41 while (cin>>a>>b) { 42 cout<<solve(b)-solve(a-1)<<endl; 43 } 44 return 0; 45 }
hdu4352 http://acm.hdu.edu.cn/showproblem.php?pid=4352
题意：求[L,R]内最长递增子序列是k的个数；
分析：知道题意后，马上map<vector<>,LL> mp[][]搞了，然后华丽丽的T掉了，
vector<int> g, g[i]表示最长递增子序列为长度为i结尾的值的最小值；
这是我对vector<>里面的值的性质没有思考，显然vector<>最多包含10个数，并且是严格递增的，
这样我们就可以直接状压存了，（1<<10）一个数值跟vector<>是一一对应的；
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #include<vector> 8 using namespace std; 9 typedef long long LL; 10 LL dp[22][1<<10][12]; 11 LL a,b; 12 int k; 13 int dig[22]; 14 int cge(int sta,int k) { 15 if (sta & (1<<k)) return sta; 16 if ((1<<k) > sta) { 17 return sta | (1<<k); 18 } 19 sta |= 1<<k; 20 for (int i = k+1; i < 10; i++) { 21 if (sta & (1<<i)) { 22 return sta ^ (1<<i); 23 } 24 } 25 } 26 int get(int k) { 27 int ret = 0; 28 for (int i = 0; i < 10; i++) if (k & (1<<i)) ret++; 29 return ret; 30 } 31 LL dfs(int pos,int sta,int limit) { 32 if (pos < 0) return get(sta) == k; 33 if (!limit && dp[pos][sta][k] != -1) 34 return dp[pos][sta][k]; 35 int last = limit ? dig[pos] : 9; 36 LL ret = 0; 37 for (int i = 0; i <= last; i++) { 38 ret += dfs(pos-1,sta || i ? cge(sta,i) : 0,limit && (i == last)); 39 } 40 if (!limit) { 41 dp[pos][sta][k] = ret; 42 } 43 return ret; 44 } 45 LL solve(LL n) { 46 int len = 0; 47 while (n) { 48 dig[len++] = n % 10; 49 n /= 10; 50 } 51 return dfs(len-1,0,1); 52 } 53 int main(){ 54 memset(dp,-1,sizeof(dp)); 55 int T,cas = 0; scanf("%d",&T); 56 while (T--) { 57 scanf("%I64d%I64d%d",&a,&b,&k); 58 printf("Case #%d: ",++cas); 59 printf("%I64d\n",solve(b) - solve(a-1)); 60 } 61 return 0; 62 }
hdu3886 http://acm.hdu.edu.cn/showproblem.php?pid=3886
题意：求[l,r]内满足题意条件的数的个数，给你一个字符串这里且称为标准串，要数值满足这个标准串（条件比较难以表述，看题）；
分析：dfs(pos,pre,loc,cc,limit,fg);
分别表示当前位置，前一位的数值，当前匹配到标准串中位置，跟标准串中匹配了几个数值，是否有限制，标记有无前导零；
有个trick，找了好久，如果直接来的话，比如// 1234 1234会输出2，因为12 34 @ 123 4 被当成不同的数了，我们可以规定如果标准串中有连续相同的字符，并且满足转移到下个字符了的话一定先转移；
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<cstdlib> 6 #include<cmath> 7 using namespace std; 8 typedef long long LL; 9 const int N = 100+10; 10 const LL Mod = 100000000; 11 char a[N],b[N]; 12 char stand[N]; 13 int flag; 14 void init(){ 15 int lena = strlen(a), lenb = strlen(b); 16 reverse(a,a+lena); 17 reverse(b,b+lenb); 18 flag = 1; 19 for (int i = 0; i < lena; i++) if (a[i] != '0') flag = 0; 20 if (lena == 1 && a[0] == '0') flag = 1; 21 int mark = 1; 22 if (!flag) 23 for (int i = 0; i < lena; i++) { 24 if (mark) { 25 if (a[i] > '0') { 26 a[i] = a[i] - 1; 27 mark = 0; 28 break; 29 } 30 else { 31 a[i] = '9'; 32 mark = 1; 33 } 34 } 35 } 36 } 37 int dig[N]; 38 int end; 39 int dp[N][10][N][2][2]; 40 41 int check(int pre,int nw,int id) { 42 if (id >= end) return 0; 43 if (pre < nw && stand[id] == '/') return 1; 44 if (pre == nw && stand[id] == '-') return 1; 45 if (pre > nw && stand[id] == '\\') return 1; 46 return 0; 47 48 } 49 int dfs(int pos,int pre,int loc,int cc,int limit,int fg) { 50 if (pos < 0) return loc == end - 1 && cc >= 2; 51 int t = 0; 52 if (cc >= 2) t = 1; 53 if (!limit && dp[pos][pre][loc][t][fg] != -1) 54 return dp[pos][pre][loc][t][fg]; 55 int last = limit ? dig[pos] : 9; 56 int ret = 0; 57 for (int i = 0; i <= last; i++) { 58 if (!fg) { 59 ret = (ret + dfs(pos-1,i,loc,i != 0,limit && (i==last),fg || i)) % Mod; 60 continue; 61 } 62 if (cc >= 2 && check(pre,i,loc+1) && stand[loc] == stand[loc+1]){ 63 ret = (ret + dfs(pos-1,i,loc+1,2,limit && (i==last),fg || i)) % Mod; 64 continue; 65 } 66 67 if (cc >= 2 && check(pre,i,loc+1) && stand[loc] != stand[loc+1]){ 68 ret = (ret + dfs(pos-1,i,loc+1,2,limit && (i==last),fg || i)) % Mod; 69 } 70 if (check(pre,i,loc)) 71 ret = (ret + dfs(pos-1,i,loc,cc+1,limit && (i==last),fg || i) ) % Mod; 72 } 73 if (!limit ) { 74 dp[pos][pre][loc][t][fg] = ret; 75 } 76 return ret; 77 } 78 int solve(char *s) { 79 int len = strlen(s); 80 for (int i = 0; i < len; i++) { 81 dig[i] = s[i] - '0'; 82 } 83 while (dig[len-1] == 0) len--; 84 // dig[len++] = 0; 85 return dfs(len-1,0,0,0,1,0); 86 } 87 int main(){ 88 while (~scanf("%s%s%s",stand,a,b)) { 89 init(); 90 memset(dp,-1,sizeof(dp)); 91 end = strlen(stand); 92 // cout<<a<<endl; 93 // cout<<b<<endl; 94 if (flag) printf("%08d\n",solve(b)); 95 else printf("%08d\n",(solve(b) - solve(a) + Mod) % Mod); 96 } 97 return 0; 98 }
cf 55D http://codeforces.com/problemset/problem/55/D
题意：求[L,R]之间能整除自己每一位的数的个数；
分析：1~9的最小公倍数为2520,同时记录下那些数出现过因为0，1不许要1<<8,cf内存大
dfs(pos,sta,mod,limit);
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 typedef long long LL; 8 const int Mod = 2520; 9 LL dp[20][1<<8][2520]; 10 11 LL a,b; 12 int dig[20]; 13 int check(int sta,int mod) { 14 for (int i = 2; i < 10; i++) { 15 if (sta & (1<<(i-2))) { 16 if (mod % i) return 0; 17 } 18 } 19 return 1; 20 } 21 LL dfs(int pos,int sta,int mod,int limit) { 22 if (pos < 0) return check(sta,mod); 23 if (!limit && dp[pos][sta][mod] != -1) return dp[pos][sta][mod]; 24 int last = limit ? dig[pos] : 9; 25 LL ret = 0; 26 for (int i = 0; i <= last; i++) { 27 int t = sta; 28 if (i >= 2) t |= 1<<(i-2); 29 ret += dfs(pos-1,t,(mod * 10 + i) % Mod,limit && (i == last)); 30 } 31 if (!limit) { 32 dp[pos][sta][mod] = ret; 33 } 34 return ret; 35 } 36 LL solve(LL n) { 37 int len = 0; 38 while (n) { 39 dig[len++] = n % 10; 40 n /= 10; 41 } 42 return dfs(len-1,0,0,1); 43 } 44 int main(){ 45 memset(dp,-1,sizeof(dp)); 46 int T; scanf("%d",&T); 47 while (T--) { 48 cin>>a>>b; 49 cout<<solve(b) - solve(a-1)<<endl; 50 } 51 return 0; 52 }
Foj 2042 http://acm.fzu.edu.cn/problem.php?pid=2042
题意：求[a,b]与[c,d]之间 xor值大于e的 sum += i ^ j;
分析：
dfs(pos,limita,limitb,limitc);
如果只是记录dp[pos]的话会TLE,所以记录dp[pos][limita][limitb][limitc];
这样的话每次solve()都要初始化；
递归版本的数位Dp记录真的很灵活，全看题目；
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #define MP make_pair 6 using namespace std; 7 const int N = 66; 8 typedef long long LL; 9 const LL Mod = 1000000007; 10 typedef pair<LL,LL> pLL; 11 pLL dp[N][2][2][2]; 12 LL a,b,c,d,e; 13 int diga[N],digb[N],digc[N]; 14 void change(int dig[],LL n,int &len) { 15 len = 0; 16 while (n) { 17 dig[len++] = n % 2; 18 n /= 2; 19 } 20 } 21 pLL dfs(int pos,int limita,int limitb,int limitc) { 22 if (pos < 0) { 23 return !limitc ? MP(1,0) : MP(0,0); 24 } 25 if (dp[pos][limita][limitb][limitc].first != -1 && dp[pos][limita][limitb][limitc].second != -1) return dp[pos][limita][limitb][limitc]; 26 int lasta = limita ? diga[pos] : 1; 27 int lastb = limitb ? digb[pos] : 1; 28 int lastc = limitc ? digc[pos] : -1; 29 pLL ret = MP(0,0); 30 for (int i = 0; i <= lasta; i++) { 31 for (int j = 0; j <= lastb; j++) { 32 int t = (i ^ j); 33 if (t >= lastc) { 34 pLL cnt = dfs(pos-1,limita && (i == lasta),limitb && (j == lastb),limitc && (t == lastc)); 35 ret.first = (ret.first + cnt.first) % Mod; 36 ret.second = ((ret.second + (1ll<<pos) * (i^j) % Mod * cnt.first % Mod) % Mod + cnt.second) % Mod; 37 } 38 } 39 } 40 dp[pos][limita][limitb][limitc] = ret; 41 42 return ret; 43 } 44 45 LL solve(LL a,LL b,LL c) { 46 for (int i = 0; i < N; i++) 47 for (int j = 0; j < 2; j++) 48 for (int k = 0; k < 2; k++) 49 for (int x = 0; x < 2; x++) dp[i][j][k][x] = MP(-1,-1); 50 51 int lena,lenb,lenc; 52 change(diga,a,lena); 53 change(digb,b,lenb); 54 change(digc,c,lenc); 55 int len = max(lena,max(lenb,lenc)); 56 while (lena < len) diga[lena++] = 0; 57 while (lenb < len) digb[lenb++] = 0; 58 while (lenc < len) digc[lenc++] = 0; 59 60 return (dfs(len-1,1,1,1).second + Mod) % Mod; 61 } 62 int main(){ 63 int T,cas = 0; scanf("%d",&T); 64 while (T--) { 65 66 printf("Case %d: ",++cas); 67 cin>>a>>b>>c>>d>>e; 68 69 cout<<((solve(b,d,e) - solve(b,c-1,e) + Mod) % Mod - solve(a-1,d,e) + solve(a-1,c-1,e) + Mod) % Mod<<endl; 70 } 71 return 0; 72 }
