Codeforces Round #396 (Div. 2)题解

A Mahmoud and Longest Uncommon Subsequence

解题思路整理：

最长不相同串，思路很简单，如果两个串相同，则肯定不存在不相同串，如果两个串不相同，则取最长串的长度~

解题代码：

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};const int N = 1e7+5;string a,b;int main(){    cin >> a>>b;    if(a==b)    {        cout << "-1"<<endl;    }    else    {        cout << max(a.size(),b.size())<<endl;;    }    return 0;   } 

B - Mahmoud and a Triangle

解题思路整理：

贪心算法：
首先排序，相邻三个数相差最小，只判断相邻三个数是否可以组成三角形，然后1-n暴力搜一遍

解题代码：

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};const int N = 1e7+5;int n,t;vi a;int main(){    cin >>n;    rep(i,1,n)    {        cin >> t;        a.push_back(t);     }    sort(a.begin(),a.end());    for(int i=0;i<n-2;i++)    {        if(a[i]+a[i+1]>a[i+2]&&a[i+1]+a[i+2]>a[i]&&a[i]+a[i+2]>a[i+1])        {            cout << "YES"<<endl;            return 0;        }    }    cout << "NO"<<endl;    return 0;   } 

C. Mahmoud and a Message 划分DP

解题思路整理：

简单dp：
我首先想到的是区间dp，但是时间复杂度为O(n^3)，所以超时，
后来看到题解为简单dp,时间复杂度为O(n^2).一个dp[]数组，从右到左根据划分遍历。

解题代码：

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};#define MOD (ll)(1e9+7)const int N = 1e7+5;bool judge(int l,int r);string s;ll n;int a[30];int re=0;ll dp[1005];ll dp1[1005];int main(){    cin >>n;    cin >> s;    rep(i,0,25)    {        cin >> a[i];    }    dp1[1]=1;    dp[0]=1;    for(int len=1;len<=n;len++)    {        dp1[len] = inf_int;        for(int i=1;i<=len;i++)        {            if(judge(i,len))            {                dp[len] += dp[i-1]%MOD;                dp[len] %= MOD;                dp1[len] = min(dp1[len],dp1[i-1]+1);            }        }    }    cout << dp[n]<<endl;    cout <<re<<endl;    cout <<dp1[n]<<endl;    return 0;   } bool judge(int l,int r){     int len = r-l+1;    int ct=a[s[l-1]-'a'];    for(int kk=l;kk<=r;kk++)    {        ct = min(ct,a[s[kk-1]-'a']);    }    if(ct>=len)    {        if(len>re)            re = len;    //  cout << l<<" "<<r<<endl;        return true;    }    else        return false;}