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CCPC2018-湖南全国邀请赛

作者:互联网

A - Easy $h$-index

思路:

这题题目说的太晦涩难懂了,英语不好直接炸掉,简单来说就是找到一个引用次数,使得至少引用这些次数的文章的数量要大于这个引用数,因为是至少,所以比如说至少引用3次,那么引用四次五次也是属于至少引用了三次的,所以由于要求的是最大的引用次数,所以我们把a从后往前加起来,直到大于当前引用次数时即为答案

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <map>
#include <unordered_set>
#include <unordered_map>

#define x first
#define y second
#define IOS ios::sync_with_stdio(false);cin.tie(0);

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 200010, M = 200010, MOD = 1000000007, INF = 0x3f3f3f3f;

int a[N];

int main()
{
    IOS;
    int n;
    while(cin >> n)
    {
        for (int i = 0; i <= n; i ++ )
            cin >> a[i];
        LL sum = 0, res = 0;
        for (int i = n; i >= 0; i -- )
        {
            sum += a[i];
            if(sum >= i)
            {
                res = i;
                break;
            }
        }
        cout << res << endl;
    }
    return 0;
}

 

B - Higher $h$-index

思路:

要使文章数最大,可知每篇文章工作一小时满足情况,由于题目说之前的文章会对现在的文章的引用加一,即多的这一个引用也算到之前的那篇文章里,所以第i篇文章当前的引用为a,多出来的引用为他之后的文章的数量,即(n - i),所以第i篇文章的总引用数为a + n - i,同上一题要>=i,所以i <= (n + a) / 2

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <map>
#include <unordered_set>
#include <unordered_map>

#define x first
#define y second
#define IOS ios::sync_with_stdio(false);cin.tie(0);

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 200010, M = 200010, MOD = 1000000007, INF = 0x3f3f3f3f;

int a[N];

int main()
{
    IOS;
    int n, a;
    while(cin >> n >> a)
    {
        cout << (n + a) / 2 << endl;
    }
    return 0;
}

F - Sorting

思路:

给出若干个三元组(a,b,c),按照(a+b)/(a+b+c)排序。首先不能直接排序,会卡精度。移项除变乘sort就行,但需要再化简一下,否则会爆long long。

对于分数有这么一个性质:

 

所以原来的按照(a[i-1]+b[i-1])*(a[i]+b[i]+c[i])<(a[i]+b[i])*(a[i-1]+b[i-1]+c[i-1])可以变为(a[i-1]+b[i-1])*c[i]<(a[i]+b[i])*c[i-1]

最后注意输出格式

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <map>
#include <unordered_set>
#include <unordered_map>

#define x first
#define y second
#define IOS ios::sync_with_stdio(false);cin.tie(0);

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 1010, M = 200010, MOD = 1000000007, INF = 0x3f3f3f3f;

struct Tuple{
    int id;
    LL a, b, c;
    bool operator< (const Tuple& t) const 
    {
        LL n1 = (a + b) * t.c, n2 = (t.a + t.b) * c;
        if(n1 == n2)
            return id < t.id;
        else
            return n1 < n2;
    }
} arr[N];

int main()
{
    IOS;
    int n;
    while(cin >> n)
    {
        for (int i = 1; i <= n; i ++ ) 
        {
            cin >> arr[i].a >> arr[i].b >> arr[i].c;
            arr[i].id = i;
        }
        sort(arr + 1, arr + 1 + n);

        for (int i = 1; i <= n; i ++ )
        {
            cout << arr[i].id;
            if(i != n)
                cout << ' ';
        }
        cout << endl;
    }
    return 0;
}

 

标签:CCPC2018,arr,int,long,引用,湖南,include,邀请赛,define
来源: https://www.cnblogs.com/yctql/p/15440951.html