https://codeforces.com/contest/1143/problem/B

題解：數(shù)位DP?

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str; struct node{}; int shu[20], dp[20];int dfs(int len, bool shangxian) {if (len == 0)return 1;if (!shangxian && dp[len]) //為什么要返回呢？可以畫(huà)圖理解當(dāng)我們搜到3XXX時(shí)，程序運(yùn)行到1XXX時(shí)就已經(jīng)把3XXX之后的搜索完了，記憶化也是這個(gè)用意.return dp[len];int res= 0, maxx = (shangxian ? shu[len] : 9);for (int i = 0; i <= maxx; i++){if(i==0)res=max(res,dfs(len - 1,shangxian && i == maxx)); //只有之前有限制現(xiàn)在的達(dá)到了上限才能構(gòu)成限制elseres=max(res,i*dfs(len - 1,shangxian && i == maxx));}return shangxian ? res: dp[len] = res; //如果有限制，那么就不能記憶化，否則記憶的是個(gè)錯(cuò)誤的數(shù). }int solve(int x) {memset(shu, 0, sizeof(shu));int k = 0;while (x){shu[++k] = x % 10; //保存a,b的數(shù)x /= 10;}return dfs(k, true); } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);cout<<solve(n)<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

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