【LGR-154-Div.4】洛谷入门赛 #15 赛后总结

整体评价

以我的水平就只能做做入门赛了，其他的比赛根本提不上劲。

Debug速度太慢导致题目没有做完。

T1 443 pts (500 pts)

考试的时候迟到了十分钟。
第一题还是很简单一个顺序结构就可解决问题

#include <iostream>
using namespace std;
int a, b, c, d;
int main() {
    cin >> a >> b >> c >> d;
    int sum = a + b + c + d;
    if(sum < 51) {
        cout << "Rabbit wins" << endl;
    }
    else {
        cout << "Rabbit lose" << endl;
    }
    return 0;
}

T2 562 pts (700 pts)

用枚举即可解决问题

#include <iostream>
#include <cstdio>
using namespace std;
int sx, sy, cx, cy, mx, my;
int fx[8] = {-1, -2, -2, -1, 1, 2, 2, 1};
int fy[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
bool check(int x,int y) {
    int sum = 0;
    for(int i = 0; i < 8; i++) {
        int nx = fx[i] + x;
        int ny = fy[i] + y;
        if((nx == sx && ny == sy) || (nx == cx && ny == cy)) sum++;
        if(sum == 2) {
            return true;
        }
    }
    return false;
}
int main() {    
    cin >> sx >> sy >> cx >> cy >> mx >> my;
    for(int i = 0; i < 8; i++) {
        int nx = fx[i] + mx;
        int ny = fy[i] + my;
        if(check(nx, ny)) {
            cout << "Yes";
            return 0;
        }
    }
    cout << "No" << endl;
    return 0;
}

T3 772 pts (900pts)

一个找规律，加上一个STL map即可解决问题

#include <iostream>
#include <map>
using namespace std;

map<int,int> M;
int n;

int main() {
    for(int i = 1; i <= 6; i += 2) {
        M[i] = i + 1;
        M[i + 1] = i;
    }
    cin >> n;
    int ans = 0;
    for(int i = 1, op; i <= n; i++) {
        cin >> op;
        if(i == n) {
            ans += 21 - M[op];
        }
        else {
            ans += 21 - M[op] - op;
        }
    }
    cout << ans << endl;
    return 0;
}

T4 681 pts (1100pts)

暴力加上一个存放Max的数组即可

#include <iostream>
#include <algorithm>
#include <limits.h>
using namespace std;
const int maxn = 1e6 + 5;
typedef long long ll;
#define int long long
int n, c;
int a[maxn], Max[maxn];

signed main() {
    cin >> n >> c;
    Max[0] = 0;
    for(int i = 1; i <= n; i++) {
        cin >> a[i];
        Max[i] = max(Max[i - 1], a[i]);
    }
    ll ansval = LLONG_MAX;
    int ansindex;
    for(int k = 0; k <= n; k++) {
        ll t = Max[k] + c * (n - k);
        if(t < ansval) {
            ansval = t;
            ansindex = k;
        }
    }
    cout << ansindex << " " << ansval << endl;
    return 0;
}

T5 852 pts (1300 pts)

Map 即可轻松解决

#include <iostream>
#include <map>
using namespace std;

int n, m;
map <int, int> score;

int main() {
    cin >> n >> m;
    int vacant = 0;
    for(int i = 1, kkksc03; i <= n; i++){
        cin >> kkksc03;
        score[kkksc03] = -1;
    }
    for(int i = 1, x, y; i <= m; i++) {
        cin >> x >> y;
        score[x] = y;
    }
    int ans = 0;
    for(auto i = score.begin(); i != score.end(); i++) {
        if(i->second < 60) ans++;
        if(i->second == -1) vacant ++;
    }
    cout << vacant << "\n" << ans << "\n";
    return 0;
}

T6 713 pts (1300 pts)

就是这道题目把我卡住了，检查了很久，结果发现是数组开小了
看来如果一道题不是爆0应该先检查数据类型，数组是不是开小了，像这种一般你的思路是对的，只不过是实现挂了

#include <iostream>
#include <vector>
using namespace std;
const int maxn = 500 * 500 + 5; //就是这里的问题,之前写成500 + 5，没有考虑到地图是二维的QAQ
int n, m, N, M;
/*
vector <int> mapH[maxn];//雄性  横向
vector <int> mapS[maxn];//ts    纵向
*/
int xF[maxn], yF[maxn];//雌性
int xM[maxn], yM[maxn];//xiong

char map[505][505];

int main(){
    cin >> n >> m >> N >> M;
    int cnt = 1;
    int cnt2 = 1;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++) {
            cin >> map[i][j];
            if(map[i][j] == 'F') {
                xF[cnt] = i;
                yF[cnt++] = j;
            }
            if(map[i][j] == 'M') {
                xM[cnt2] = i;
                yM[cnt2++] = j;
            }
        }
    int sum = 0;
    for(int i = 1; i <= cnt; i++) {
        int Run = 0;
        int x, y;
        x = xF[i];
        y = yF[i];
        while(map[x - 1][y] != 'M' && x - 1 >= 1) 
            x--;
        if(x == 1) Run++;
        x = xF[i];
        y = yF[i];
        while(map[x + 1][y] != 'M' && x + 1 <= n ) 
            x++;
        if(x == n) Run++;
        x = xF[i];
        y = yF[i];
        while(map[x][y - 1] != 'M' && y - 1 >= 1) 
            y--;
        if(y == 1) Run++;
        x = xF[i];
        y = yF[i];
        while(map[x][y + 1] != 'M' && y + 1 <= m) 
            y++;
        if(y == m) Run++;
        x = xF[i];
        y = yF[i];
        if(Run >= 3) 
            sum++;
    }
    for(int i = 1; i <= cnt2; i++) {
        int Run = 0;
        int x, y;
        x = xM[i];
        y = yM[i];
        while(map[x - 1][y] != 'F' && x - 1 >= 1) 
            x--;
        if(x == 1) Run++;
        x = xM[i];
        y = yM[i];
        while(map[x + 1][y] != 'F' && x + 1 <= n) 
            x++;
        if(x == n) Run++;
        x = xM[i];
        y = yM[i];
        while(map[x][y - 1] != 'F' && y - 1 >= 1) 
            y--;
        if(y == 1) Run++;
        x = xM[i];
        y = yM[i];
        while(map[x][y + 1] != 'F' && y + 1 <= m) 
            y++;
        if(y == m) Run++;
        x = xM[i];
        y = yM[i];
        if(Run >= 3) 
            sum++;
    }
    cout << sum << endl;
    return 0;
}

T7 694 pts (1700 pts)

也是STL map 解决问题

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
using namespace std;

map<string, string> name;

int main() {
    int n, m, l;
    cin >> n >> m >> l;
    for(int i = 1; i <= n; i++) {
        string Input;
        cin >> Input;
        name[Input] = Input;
    }
    for(int i = 1; i <= m; i++) {
        string Input;
        cin >> Input;
        name.erase(Input);
    }
    for(int i = 1; i <= l; i++) {
        string Input;   
        cin >> Input;
        name[Input] = Input;
    }
    vector <string> ls;
    for(auto i = name.begin(); i != name.end(); i++) {
        ls.push_back(i->second);
    }
    sort(ls.begin(), ls.end());
    for(int i = 0; i < ls.size(); i++){
        cout << ls[i] << endl;
    }
    return 0;
}

T8 0 pts (2000 pts)

这道题在世界这么短的情况下做是没有意义的，于是我就跳到了T9

T9 750pts (1500pts)

这道题只看了第一个程序，爆int了

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n, k;
    cin >> n >> k;
    if (k == n || k == 0) {
        cout << "0 0" << endl;
        return 0;
    }
    cout << "1 ";
    if (k * 3 <= n)//就是这里爆掉了
        cout << 2 * k << endl;
    else
        cout << n - k << endl;
    return 0;
}

总结

• debug能力有待提高