2020牛客暑期多校训练营（第五场）

F DPS

 

(坑点) 如果使用int会爆，使用double会缺失精度，因此使用long long。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[110];
int main()
{
    memset(a,0,sizeof(a));
    ll n,i,ma=-1;
    scanf("%lld",&n);
    for(i=0; i<n; i++)
    {
        scanf("%lld",&a[i]);
        ma=max(ma,a[i]);
    }
    for(i=0; i<n; i++)
    {
        printf("+");
        ll s=ceil(50*a[i]*1.0/ma);
        for(ll j=0; j<s; j++)
        {
            printf("-");
        }
        printf("+\n");
        printf("|");
        for(ll j=0; j<s-1; j++)
        {
            printf(" ");
        }
        if(a[i]==ma)
        {
            printf("*");
        }
        else if(a[i]!=0)
        {
            printf(" ");
        }
        printf("|");
        printf("%d\n",a[i]);
        printf("+");
        for(ll j=0; j<s; j++)
        {
            printf("-");
        }
        printf("+\n");
    }
}

I Hard Math Problem

来自pdf

题意：

　　• 有一个无穷大的二维网格，每个格子可以是1、2或者3，每个1旁边要有一个2和3，要使机器的占比 最大。
做法：

　　• (i+j)%3 =0 交错2和3
　　• 答案是2/3 • 因为这样每个1旁边恰好有一个2和3，而任意两个2和3不相邻。

　　• 可以计算出这是上限。

代码略

E Bogo Sort

关键在于，大数板子(Java和python选手除外)和如何找环

大数板子

int max(int a, int b) { return a>b?a:b; }
struct bign {
    int len, s[numlen];
    bign() {
        memset(s, 0, sizeof(s));
        len = 1;
    }
    bign(int num) { *this = num; }
    bign(const char *num) { *this = num; }
    bign operator = (const int num) {
        char s[numlen];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bign operator = (const char *num) {
        len = strlen(num);
        while(len > 1 && num[0] == '0') num++, len--;
        for(int i = 0;i < len; i++) s[i] = num[len-i-1] - '0';
        return *this;
    }
  
    void deal() {
        while(len > 1 && !s[len-1]) len--;
    }
  
    bign operator + (const bign &a) const {
        bign ret;
        ret.len = 0;
        int top = max(len, a.len) , add = 0;
        for(int i = 0;add || i < top; i++) {
            int now = add;
            if(i < len) now += s[i];
            if(i < a.len)   now += a.s[i];
            ret.s[ret.len++] = now%10;
            add = now/10;
        }
        return ret;
    }
    bign operator - (const bign &a) const {
        bign ret;
        ret.len = 0;
        int cal = 0;
        for(int i = 0;i < len; i++) {
            int now = s[i] - cal;
            if(i < a.len)   now -= a.s[i];
            if(now >= 0)    cal = 0;
            else {
                cal = 1; now += 10;
            }
            ret.s[ret.len++] = now;
        }
        ret.deal();
        return ret;
    }
    bign operator * (const bign &a) const {
        bign ret;
        ret.len = len + a.len;
        for(int i = 0;i < len; i++) {
            for(int j = 0;j < a.len; j++)
                ret.s[i+j] += s[i]*a.s[j];
        }
        for(int i = 0;i < ret.len; i++) {
            ret.s[i+1] += ret.s[i]/10;
            ret.s[i] %= 10;
        }
        ret.deal();
        return ret;
    }
  
    //乘以小数，直接乘快点
    bign operator * (const int num) {
        bign ret;
        ret.len = 0;
        int bb = 0;
        for(int i = 0;i < len; i++) {
            int now = bb + s[i]*num;
            ret.s[ret.len++] = now%10;
            bb = now/10;
        }
        while(bb) {
            ret.s[ret.len++] = bb % 10;
            bb /= 10;
        }
        ret.deal();
        return ret;
    }
  
    bign operator / (const bign &a) const {
        bign ret, cur = 0;
        ret.len = len;
        for(int i = len-1;i >= 0; i--) {
            cur = cur*10;
            cur.s[0] = s[i];
            while(cur >= a) {
                cur -= a;
                ret.s[i]++;
            }
        }
        ret.deal();
        return ret;
    }
  
    bign operator % (const bign &a) const {
        bign b = *this / a;
        return *this - b*a;
    }
  
    bign operator += (const bign &a) { *this = *this + a; return *this; }
    bign operator -= (const bign &a) { *this = *this - a; return *this; }
    bign operator *= (const bign &a) { *this = *this * a; return *this; }
    bign operator /= (const bign &a) { *this = *this / a; return *this; }
    bign operator %= (const bign &a) { *this = *this % a; return *this; }
  
    bool operator < (const bign &a) const {
        if(len != a.len)    return len < a.len;
        for(int i = len-1;i >= 0; i--) if(s[i] != a.s[i])
            return s[i] < a.s[i];
        return false;
    }
    bool operator > (const bign &a) const  { return a < *this; }
    bool operator <= (const bign &a) const { return !(*this > a); }
    bool operator >= (const bign &a) const { return !(*this < a); }
    bool operator == (const bign &a) const { return !(*this > a || *this < a); }
    bool operator != (const bign &a) const { return *this > a || *this < a; }
  
    string str() const {
        string ret = "";
        for(int i = 0;i < len; i++) ret = char(s[i] + '0') + ret;
        return ret;
    }
};
istream& operator >> (istream &in, bign &x) {
    string s;
    in >> s;
    x = s.c_str();
    return in;
}
ostream& operator << (ostream &out, const bign &x) {
    out << x.str();
    return out;
}
// 大数开平方
bign Sqrt(bign x) {
    int a[numlen/2];
    int top = 0;
    for(int i = 0;i < x.len; i += 2) {
        if(i == x.len-1) {
            a[top++] = x.s[i];
        }
        else
            a[top++] = x.s[i] + x.s[i+1]*10;
    }
    bign ret = (int)sqrt((double)a[top-1]);
    int xx = (int)sqrt((double)a[top-1]);
    bign pre = a[top-1] - xx*xx;
    bign cc;
    for(int i = top-2;i >= 0; i--) {
        pre = pre*100 + a[i];
        cc = ret*20;
        for(int j = 9;j >= 0; j--) {
            bign now = (cc + j)*j;
            if(now <= pre) {
                ret = ret*10 + j;
                pre -= now;
                break;
            }
        }
    }
    return ret;
}

找环

for(int i=1; i<=n; i++)
    {
        if(!vis[i])
        {
            num.clear();
            int tmp=i;
            while(!vis[tmp])
            {
                vis[tmp]=1;
                num.push_back(tmp);
                tmp=a[tmp];
            }
            f();
        }
    }

vis数组用来存放每个数是否被查询的状态，当未被查询时，进入循环，将当前这个数push_back到num容器中去。i从1到n为止。