CodeForces Round # 526(div 2)

A：The Fair Nut and Elevator

题意：问电梯放在那一层的花费最小。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
int a[N];
int main(){
    int ans = inf;
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
    for(int i = 1; i <= n; ++i){
        int tmp = 0;
        for(int j = 1; j <= n; ++j){
            tmp += a[j] * 2 * (abs(i-j)+abs(j-1)+abs(1-i));
        }
        ans = min(ans, tmp);
    }
    cout << ans << endl;
    return 0;
}

B：Kvass and the Fair Nut

题意：有n桶水， 问一共需要拿走k滴水，要求拿走之后n桶水最小的那个最大。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
LL a[N];
int main(){
    int n; LL s;
    scanf("%d%lld", &n, &s);
    for(int i = 1; i <= n; ++i)
        scanf("%lld", &a[i]);
    sort(a+1, a+1+n);
    LL tmp = 0;
    for(int i = 2; i <= n; ++i){
        tmp += a[i] - a[1];
    }
    LL need = s - tmp;
    if(need <= 0){
        printf("%lld
", a[1]);
    }
    else {
        need = (need + n-1) /n;
        a[1] -= need;
        if(a[1] < 0) puts("-1");
        else printf("%lld
", a[1]);
    }
    return 0;
}

C：The Fair Nut and String

题解：

定义 ok 为 后面加了'a' 之后的合法序列数目， _ok为 后面加了 'b'之后可以变成ok的序列数目。

那么 每次遇到一个'a'就可以  ans += ok， _ok += ok,

然后遇到一个'b'就可以使得 ok += _ok, _ok = 0;

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
char s[N];
LL ans = 0, ok = 0, _ok = 0;
int main(){
    scanf("%s", s+1);
    int n = strlen(s+1);
    for(int i = 1; i <= n; ++i){
        if(s[i] == 'a'){
            ans += ok + 1;
            ans %= mod;
            _ok += ok + 1;
            _ok %= mod;
        }
        if(s[i] == 'b'){
            ok += _ok;
            _ok = 0;
            ok %= mod;
        }
    }
    cout << ans << endl;
    return 0;
}

D：The Fair Nut and the Best Path

题解

E：The Fair Nut and Strings

题意：问在这[s,t]的串中选k个串，不同前缀的数目最大是多少。

题解：

定义 ok  为 第i位的不同前缀和的数目。

那么当i->i+1的时候， 如果我们不考虑 s, t的限制的话 

ok -> ok * 2

限制又有了st的限制，如果 s[i+1] == 'b' 那么就说明 可以选择的方案数--， t[i+1] == 'a'， 可以选择的方案数--，

注意不要爆LL就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 5e5 + 100;
char s[N], t[N];
int main(){
    LL n, k;
    LL ans = 0, ok = 1;
    scanf("%lld%lld", &n, &k);
    scanf("%s%s", s, t);
    for(int i = 0; i < n; ++i){
        ok = ok * 2;
        if(s[i] == 'b') ok--;
        if(t[i] == 'a') ok--;
        if(ok > mod) ok = mod;
        ans += min(ok, k);
    }
    cout << ans << endl;
    return 0;
}

F：Max Mex

题解