数位dp题集
题集见大佬博客
不要62
入门题，检验刚才自己有没有看懂
注意一些细节。
的确挺套路的
#include<bits/stdc++.h> #define REP(i, a, b) for(register int i = (a); i < (b); i++) #define _for(i, a, b) for(register int i = (a); i <= (b); i++) using namespace std; const int MAXN = 15; int a[MAXN], dp[MAXN][MAXN][MAXN], len; //dp数组记录除了lead和limit以外其他的东西 int dfs(int pos, int pre, int lead, int sum, int limit) { if(pos > len) return sum; if(dp[pos][pre][sum] != -1 && !limit && !lead) return dp[pos][pre][sum]; //记忆化的时候记得limit和lead int l = limit ? a[len-pos+1] : 9, res = 0; //注意是倒序存的，所以是len-pos+1 _for(i, 0, l) { if(!i && lead) res += dfs(pos + 1, pre, lead, sum, limit && (i == l)); //先看是不是前导0 else if(i && lead) res += dfs(pos + 1, i, 0, sum | (i == 4), limit && (i == l)); //看是不是第一位 else res += dfs(pos + 1, i, 0, sum | (i == 4) | (pre == 6 && i == 2), limit && (i == l)); //正式处理 } return (!limit && !lead) ? dp[pos][pre][sum] = res : res; //记忆化的时候记得limit和lead } int part(int x) { if(x < 0) return 0; //0的处理 memset(dp, -1, sizeof(dp)); len = 0; int t = x; for(; x; x /= 10) a[++len] = x % 10; return t - dfs(1, 0, 1, 0, 1); } int main() { int n, m; while(scanf("%d%d", &n, &m)) { if(n == 0 && m == 0) break; printf("%d\n", part(m) - part(n - 1)); } return 0; }
P2657 [SCOI2009]windy数
继续套路
#include<bits/stdc++.h> #define REP(i, a, b) for(register int i = (a); i < (b); i++) #define _for(i, a, b) for(register int i = (a); i <= (b); i++) using namespace std; const int MAXN = 15; int a[MAXN], dp[MAXN][MAXN][MAXN], len; int dfs(int pos, int pre, int ans, int lead, int limit) { if(pos > len) return ans; if(dp[pos][pre][ans] != -1 && (!limit && !lead)) return dp[pos][pre][ans]; int l = limit ? a[len-pos+1] : 9, res = 0; _for(i, 0, l) { if(!i && lead) res += dfs(pos + 1, pre, ans, lead, limit & (i == l)); else if(i && lead) res += dfs(pos + 1, i, ans, 0, limit & (i == l)); else res += dfs(pos + 1, i, ans & (abs(i - pre) >= 2), 0, limit & (i == l)); } return (!limit && !lead) ? dp[pos][pre][ans] = res : res; } int part(int x) { if(x == 0) return 1; memset(dp, -1, sizeof(dp)); len = 0; for(; x; x /= 10) a[++len] = x % 10; return dfs(1, 0, 1, 1, 1); } int main() { int a, b; scanf("%d%d", &a, &b); printf("%d\n", part(b) - part(a - 1)); return 0; }
P2602 [ZJOI2010]数字计数
第一次这么轻松做出紫题
一样套模板，爽啊
#include<bits/stdc++.h> #define REP(i, a, b) for(register int i = (a); i < (b); i++) #define _for(i, a, b) for(register int i = (a); i <= (b); i++) using namespace std; typedef long long ll; const int MAXN = 20; const int MAXM = 1e6 + 10; ll dp[MAXN][MAXN]; int a[MAXN], len; ll dfs(int pos, int ans, int lead, int limit, int key) { if(pos > len) return ans; if(dp[pos][ans] != -1 && (!lead && !limit)) return dp[pos][ans]; int l = limit ? a[len-pos+1] : 9; ll res = 0; _for(i, 0, l) { if(!i && lead) res += dfs(pos + 1, ans, lead, limit && (i == l), key); else res += dfs(pos + 1, ans + (i == key), 0, limit && (i == l), key); } return (!lead && !limit) ? dp[pos][ans] = res : res; } ll part(ll x, int i) { memset(dp, -1, sizeof(dp)); len = 0; for(; x > 0; x /= 10) a[++len] = x % 10; return dfs(1, 0, 1, 1, i); } inline ll work(ll a, ll b, int i) { return a ? part(b, i) - part(a - 1, i) : part(b, i) - part(a, i) + (i == 0); } int main() { ll a, b; scanf("%lld%lld", &a, &b); _for(i, 0, 9) printf("%lld%c", work(a, b, i), i == 9 ? '\n' : ' '); return 0; }
P3413 SAC#1 - 萌数
注意数字很大，要用字符串存储
然后就没啥了，又独立做出紫题
#include<bits/stdc++.h> #define add(a, b) a = (a + b) % mod #define REP(i, a, b) for(register int i = (a); i < (b); i++) #define _for(i, a, b) for(register int i = (a); i <= (b); i++) using namespace std; const int MAXN = 12; const int MAXM = 1e3 + 10; const int mod = 1e9 + 7; int dp[MAXM][MAXN][MAXN][2]; int a[MAXM], len; char s1[MAXM], s2[MAXM]; int dfs(int pos, int pre, int ppre, int ans, int lead, int limit) { if(pos > len) return ans; if(dp[pos][pre][ppre][ans] != -1 && (!lead && !limit)) return dp[pos][pre][ppre][ans]; int l = limit ? a[len-pos+1] : 9; int res = 0; _for(i, 0, l) { if(!i && lead) add(res, dfs(pos + 1, pre, ppre, ans, lead, limit && (i == l))); else if(i && lead) add(res, dfs(pos + 1, i, pre, ans, 0, limit && (i == l))); else add(res, dfs(pos + 1, i, pre, ans | (i == pre) | (i == ppre), 0, limit && (i == l))); } return (!lead && !limit) ? dp[pos][pre][ppre][ans] = res : res; } int part(char* s) { memset(dp, -1, sizeof(dp)); len = strlen(s + 1); _for(i, 1, len) a[i] = s[len - i + 1] - '0'; return dfs(1, -1, -1, 0, 1, 1); } int judge(char* s) { _for(i, 2, strlen(s + 1)) if(s[i] == s[i-2] || s[i] == s[i-1]) return 1; return 0; } int main() { scanf("%s%s", s1 + 1, s2 + 1); if(s1[1] == 0 && strlen(s1 + 1) == 1) printf("%d\n", (part(s2) - part(s1) + mod) % mod); else printf("%d\n", (part(s2) - part(s1) + mod + judge(s1)) % mod); return 0; }
P4127 [AHOI2009]同类分布
一开始想的时候当前的数肯定是不能记到答案里的
当然如果知道模数，一直取模就可以限制范围，就很好了
但问题是我们并不知道模数，这就比较尴尬了
就一直卡在这
然而正解非常暴力，但却是对的。
既然不知道模数，那就枚举模数
最后判断一下数字和是不是当前模数且能否整除就好了。
我为什么没想到呢？
注意数位dp要开long long
#include<bits/stdc++.h> #define REP(i, a, b) for(register int i = (a); i < (b); i++) #define _for(i, a, b) for(register int i = (a); i <= (b); i++) using namespace std; typedef long long ll; const int MAXN = 20; ll dp[MAXN][MAXN*10][MAXN*10]; int a[MAXN], len, mod; ll dfs(int pos, int num, int state, int lead, int limit) { if(pos > len) return num == mod && state == 0; if(dp[pos][num][state] != -1 && (!lead && !limit)) return dp[pos][num][state]; int l = limit ? a[len-pos+1] : 9; ll res = 0; _for(i, 0, l) { if(!i && lead) res += dfs(pos + 1, num, state, lead, limit && (i == l)); else if(i && lead) res += dfs(pos + 1, i, i % mod, 0, limit && (i == l)); else res += dfs(pos + 1, num + i, (state * 10 + i) % mod, 0, limit && (i == l)); } return (!lead && !limit) ? dp[pos][num][state] = res : res; } ll part(ll x) { len = 0; for(; x; x /= 10) a[++len] = x % 10; ll res = 0; for(mod = 1; mod <= len * 9; mod++) { memset(dp, -1, sizeof(dp)); res += dfs(1, 0, 0, 1, 1); } return res; } int main() { ll a, b; scanf("%lld%lld", &a, &b); printf("%lld\n", !a ? part(b) - part(a) : part(b) - part(a - 1)); return 0; }
P4317 花神的数论题
用二进制统计就好了
不过很奇怪的是一开始我数组开的大小是50
因为用程序输出1e15最多的位数是47
但是交上去会WA一个点
改成51就过了
以后只要空间剩余多，就多开一些吧，程序中有些神奇的地方可能会比理论上最大值超出一些
#include<bits/stdc++.h> #define mul(a, b) a = (a * b) % mod #define REP(i, a, b) for(register int i = (a); i < (b); i++) #define _for(i, a, b) for(register int i = (a); i <= (b); i++) using namespace std; typedef long long ll; const int MAXN = 51; const int mod = 10000007; ll dp[MAXN][MAXN]; int a[MAXN], len; ll dfs(int pos, int ans, int lead, int limit) { if(pos > len) return max(1, ans); if(dp[pos][ans] != -1 && (!limit && !lead)) return dp[pos][ans]; int l = limit ? a[len-pos+1] : 1; ll res = 1; _for(i, 0, l) { if(!i && lead) mul(res, dfs(pos + 1, ans, lead, limit && (i == l))); else mul(res, dfs(pos + 1, ans + i, 0, limit && (i == l))); } return (!limit && !lead) ? dp[pos][ans] = res : res; } ll part(ll x) { len = 0; for(; x; x >>= 1) a[++len] = x & 1; memset(dp, -1, sizeof(dp)); return dfs(1, 0, 1, 1); } int main() { ll a; scanf("%lld", &a); printf("%lld\n", part(a)); return 0; }
总结
感觉都是套路，掌握套路就好了
相信自己已经掌握了数位dp
