第七届福建省大学生程序设计竞赛

训练赛 ——第七届福建省大学生程序设计竞赛

rank 68/250

 

E - Card Game (Third Edition)

一号签到题

solved by mwh 16：05

=#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#pragma GCC optimize(2)

#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second

using namespace std;
typedef long long ll;
typedef pair<int, int > PII;

const int N = 1e4 + 5, mod = 1e9 + 9, INF = 0x3f3f3f3f;
int t, n;
int a[N], b[N];

int main()
{
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
//    cin.tie(0);
//    cout.tie(0);
//    ios::sync_with_stdio(0);
    cin >> t;
    for (int k = 1; k <= t; ++k) {
        ll ans = 0;
        cin >> n;
        for (int i = 1; i <= n; ++i)
            scanf("%d", &a[i]);
        for (int i = 1; i <= n; ++i)
            scanf("%d", &b[i]);
        for (int i = 1; i <= n; ++i) {
            if (a[i] + b[i] > 10)
                ans ++;
        }
        printf("Case %d: %d\n", k, ans);
    }
}

 

C - Card Game (First Edition)

二号签到题

solved  40：57（-1）

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <set>
#pragma GCC optimize(2)

#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second

using namespace std;
typedef long long ll;
typedef pair<int, int > PII;

const int N = 1e6 + 5, mod = 1e9 + 9, INF = 0x3f3f3f3f;
int t, n;
int a[N], b[N];

int main()
{
    cin >> t;
    for (int k = 1; k <= t; ++k) {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
        }
        printf("Case %d: %.2lf\n", k, 1.0 * n / 2);
    } 
}

 

D - Card Game

四号签到题

solved by mwh 01:55:25

可以发现只需枚举第一个数组中的每个数，把第二个数组中比改数小的数字个数加起来，这个和和除以n就可

可以直接用二分搜索，时间复杂度O（nlogn）

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#pragma GCC optimize(2)

#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second

using namespace std;
typedef long long ll;
typedef pair<int, int > PII;

const int N = 1e6 + 5, mod = 1e9 + 9, INF = 0x3f3f3f3f;
int t, n;
int a[N], b[N];
double ans = 0;

int main()
{
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
//    cin.tie(0);
//    cout.tie(0);
//    ios::sync_with_stdio(0);
    cin >> t;
    for (int k = 1; k <= t; ++k) {
        cin >> n;
        for (int i = 0; i < n; ++i)
            scanf("%d", &a[i]);
        for (int i = 0; i < n; ++i)
            scanf("%d", &b[i]);
        sort(a, a + n);
        sort(b, b + n);
        ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += (lower_bound(b, b + n, a[i]) - b);
        }
        printf("Case %d: %.2lf\n", k, ans / n);
    }
    
}

 

G - Cutting Game

三号签到题

直接1、2、4、8.......就可

题目非常简单，但是不知道为什么原因一直WA，这题也直接导致了我队前期节奏爆炸，导致罚时发炸

后来队友对着这题自闭自闭一个多小时后把long long改成int就过了

solved by rst 01:18:38(-3)

#include <stdio.h>
int main()
{
    int t, d;
    scanf("%d", &t);
    int k = 0;
    while(t--)
    {
        int ans = 0;
        scanf("%d", &d);
        while(d)
        {
            d /= 2;
            ans++;
        }
        printf("Case %d: %d\n", ++k, ans);
    }
}

 

J - X

显然就是一道floyd的题目

solved by mwh 02:59:35(-3)

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>

#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second

using namespace std;
typedef long long ll;
typedef pair<int, int > PII;

const int N = 110, mod = 1e9 + 9, INF = 0x3f3f3f3f;
int t, n, m, ans;
int vis[N][N];
int d[N][N];
map<PII, int> q;

int main()
{
    cin >> t;
    for (int k = 1; k <= t; ++k) {
        ans = 0;
        q.clear();
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j) {
                d[i][j] = INF;
                d[j][i] = INF;
                vis[i][j] = 0;
                vis[j][i] = 0;
            }
        for (int i = 1; i <= m; ++i) {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            if (vis[x][y] != 0) {
                if (vis[x][y] <= z) {
                    ans++;
                } else {
                    vis[x][y] = z;
                    vis[y][x] = z;
                    d[x][y] = min(d[x][y], z);
                    d[y][x] = min(d[y][x], z);
                    ans++;
                }
            } else {
                vis[x][y] = z;
                vis[y][x] = z;
                d[x][y] = min(d[x][y], z);
                d[y][x] = min(d[y][x], z);
            }
        }
        
        for (int p = 1; p <= n; ++p) {
            for (int i = 1; i <= n; ++i) {
                for (int j = 1; j <= n; ++j) {
                    if (d[i][p] + d[p][j] <= d[i][j]) {
                        d[i][j] = d[i][p] + d[p][j];
                        if (vis[i][j] != 0) {
                            ans++;
                            vis[i][j] = 0;
                            vis[j][i] = 0;
                        }
                    }
                }
            }
        }
        printf("Case %d: %d\n", k, ans);
    }
}

 

F - The Bigger the Better

笨比mwh自闭一个多小时，赛后问nb室友xwd发现我好像想的有点多？

unsolved by mwh

 

I - Two Triangles

大概是一道计算几何？反正我看都没看，两个队友对着他开始了模拟，看了一眼代码人都傻了

unsolved by czw&rst