Educational Codeforces Round 57

2018.12.28  22：30

看着CF升高的曲线，摸了摸自己的头发，我以为我变强了，直到这一场Edu搞醒了我。。

从即将进入2018年末开始，开启自闭场集合，以纪念(dian)那些丢掉的头发

留坑睡觉。。明天看题解再补

A.Find Divisible

题意：输出[l,r]中满足x|y的x,y，保证有解

思路：直接输出x, 2x即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int main() {
    int t;
    scanf("%d", &t);
    while(t--){
        ll x, y;
        scanf("%lld %lld", &x, &y);
        printf("%lld %lld\n", x,*x);
    }
    return ;
}

B.Substring Removal

题意：删除一个子串，使得剩下的串只有一种字符，问你方案数

思路：分a[1]是否==a[n]两种情况讨论，

等于的时候根据要删除的子串区间端点范围决定方案数

不等于的时候相当于一个端点固定

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = ;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);
char a[maxn];
int main() {
    int n;
    scanf("%d", &n);
    getchar();
    scanf("%s", a+);
    ll ans = ;
    if(a[]==a[n]){
        char c = a[];
        int l , r;
        for(int i = ; i <= n; i++){
            if(a[i]!=c){l=i;break;}
        }
        for(int i = n; i >= ; i--){
            if(a[i]!=c){r=i;break;}
        }
        l = l;
        r = n-r+;
        ans = 1ll*l*r;
        ans%=mod;
    }
    else{
        int l,r;
        char c = a[];
        for(int i = ; i <= n; i++){
            if(a[i]!=c){l=i;break;}
        }
        c = a[n];
        for(int i = n; i >= ; i--){
            if(a[i]!=c){r=i;break;}
        }
        ans = n-r;
        ans += l;
        ans%=mod;
    }
    printf("%lld", ans);
    return ;
}

C.Polygon for the Angle

题意：问你一个角度最少存在于正几边形中

思路：看代码。。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = ;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int main() {
    int t;
    scanf("%d", &t);
    while(t--){
        int ag;
        scanf("%d", &ag);
        int g = __gcd(ag,);
        int n = /g;
        int m = ag/g;
        while(m>n-){n*=;m*=;}
        printf("%d\n", n);
    }
    return ;
}

D.Easy Problem

题意：给你一个字符串，每个字符都有权重，问你删掉多少权重和的字符，使得剩下的字符没有"hard"子序列，并且这个权重和最小

思路：dp[i][j]为前i个字符最多拥有j状态的子序列需要删除的最小权重

其中 j=0代表空，j=1代表"h"，j=2代表"ha"，j=3代表"har"

转移方程在代码里

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = ;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);
char s[maxn];
ll dp[maxn][];
//dp[i][j]表示前i个字符最多有prej作为子序列要删多少
int a[maxn];
int n;
int main() {
    scanf("%d", &n);
    getchar();
    scanf("%s", s+);
    for(int i = ; i <= n; i++){
        scanf("%d", &a[i]);
    }
    mem(dp,);
    //dp[0][0] = 0;
    for(int i = ; i <= n; i++){
        for(int j = ; j < ; j++){
            dp[i][j] = dp[i-][j];
        }
        if(s[i]=='h')dp[i][]+=a[i];
        if(s[i]=='a')dp[i][]=min(dp[i-][],dp[i][]+a[i]);
        if(s[i]=='r')dp[i][]=min(dp[i-][],dp[i][]+a[i]);
        if(s[i]=='d')dp[i][]=min(dp[i-][],dp[i][]+a[i]);
    }
    ll ans = 0x7f7f7f7f7f7f7f7f;
    for(int i = ; i <=  ;i++){
        ans = min(ans, dp[n][i]);
    }printf("%lld", ans);
    return ;
}