浙江省高职院校联合训练（一）

题目链接：http://acm.hdu.edu.cn/webcontest/contest_show.php?cid=13137

A .Problem A(ZOJ1048普通签到题)

sol：不解释

  • 水题 
    #include "cstdio"
using namespace std;
int main() {
    double sum = 0, k;
    for (int i = 0; i < 12; i++) {
        scanf("%lf", &k);
        sum += k;
    }
    printf("$%.2lf\n", sum / 12);
    return 0;
}
    View Code

     

B .Problem B(POJ2632复杂模拟题)

sol：模拟题解法不一，做着相当恶心人。把昨天的代码理了理重写了一份

  •  恶心人的模拟题 
    #include "cstdio"
#include "string"
#include "iostream"
#include "cstring"
using namespace std;
const int MAXN = 105;
struct Robot {
    int x, y, d;
} rbt[MAXN];
int n, m;
string ans;
int mp[MAXN][MAXN];
int dir[4][2] = {1, 0, 0, -1, -1, 0, 0, 1};
// 将方向表示成int,方便转向的时候加减 
int getDir(char c) {
    switch (c) {
        case 'N' : return 0;
        case 'W' : return 1;
        case 'S' : return 2;
        case 'E' : return 3;
    }
}
// 将int转成string 
string itoa(int k) {
    if (k == 100) {
        return "100";
    }
    string s = "";
    if (k >= 10) {
        s += k / 10 + '0';
        s += k % 10 + '0';
    } else {
        s += k + '0';
    }
    return s;
}
// 检查是否越界或者撞上其他机器人 
bool check(int k, int x, int y) {
    if (x <= 0 || y <= 0 || x > n || y > m) {
        ans = "Robot ";
        ans += itoa(k);
        ans += " crashes into the wall";
        return true;
    }
    if (mp[x][y]) {
        ans = "Robot ";
        ans += itoa(k);
        ans += " crashes into robot ";
        ans += itoa(mp[x][y]);
        return true;
    }
    return false;
}
// 移动机器人 
void move(int k, int d) {
    int xx = dir[rbt[k].d][0], yy = dir[rbt[k].d][1];
    int x = rbt[k].x, y = rbt[k].y;
    mp[x][y] = 0;
    for (int i = 0; i < d; i++) {
        x += xx;
        y += yy;
        if (check(k, x, y)) return;
    }
    mp[x][y] = k;
    rbt[k].x = x;
    rbt[k].y = y;
}
// 给机器人转向 
void changeDir(int k, int d) {
    rbt[k].d = (rbt[k].d + d) % 4;
    if (rbt[k].d < 0) rbt[k].d += 4;
}
int main() {
    int t, k1, k2;
    scanf("%d", &t);
    while (t--) {
         scanf("%d%d%d%d", &m, &n, &k1, &k2);
        for (int i = 1; i <= k1; i++) {
            int x, y; char c;
            scanf("%d %d %c", &y, &x, &c);
            rbt[i] = {x, y, getDir(c)};
            mp[x][y] = i;
        }
        ans = "OK";
        for (int i = 1; i <= k2; i++) {
            int id, k; char op;
            scanf("%d %c %d", &id, &op, &k);
            if (ans[0] != 'O') continue;
            if (op == 'F') move(id, k);
            else if (op == 'L') changeDir(id, k % 4);
            else changeDir(id, -k % 4);
        }
        cout << ans << endl;
        memset(mp, 0, sizeof(mp));
    }
    return 0;
}
    View Code

     

E .Problem E(ZOJ2371递推、组合数学)

sol：先把n减1，然后从n的二进制低位往高位遍历，如果((1 << k) & n) == 1，就输出数组中的第k + 1个数。注意一下格式并且特判n - 1之后是0的情况还有高精度。

ps:JAVA自带高精度类，所以写大数首选JAVA

  • 高精度，数学 
    import java.math.BigInteger;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong() - 1;
        while (n != -1) {
            if (n == 0) {
                System.out.println("{ }");
                n = sc.nextLong() - 1;
                continue;
            }
            BigInteger pw = BigInteger.ONE;
            System.out.print("{ ");
            while (n != 1) {
                if ((n & 1) == 1) {
                    System.out.print(pw + ", ");
                }
                pw = pw.multiply(new BigInteger("3"));
                n >>= 1;
            }
            System.out.println(pw + " }");
            n = sc.nextLong() - 1;
        }
    }

}
    View Code

     

F .Problem F(POJ1064二分)

sol：二分答案，判断答案小了或是大了。

ps:用double会造成精度丢失

  • 简单二分 
    #include "cstdio"
using namespace std;
const int MAXN = 10005;
int n, m;
int cable[MAXN];
bool check(int mid) {
    int sum = 0;
    for (int i = 1; i <= n; i++) 
        sum += cable[i] / mid;
    return sum >= m;
}
int main() {
    double k;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%lf", &k);
        cable[i] = k * 100 + 0.1;
    }
    // 我们要的答案一定在0到1e9; 
    int l = 0, r = 1e9;
    while (r - l > 1) {
        int mid = l + r >> 1;
        if (check(mid)) l = mid;
        else r = mid;
    }
    printf("%d.%02d\n", l / 100, l % 100);
    return 0;
}
    View Code

    比赛时间不够，题目都来不及读。看完题发现还是挺简单的

H .Problem H(ZOJ1195简单模拟)

sol：记录每台机器状态和电流，每次转换修改sum，然后和mx比较。

ps: Output a blank line after each test case.

  • 水题 
    #include "cstdio"
#include "cstring"
#include "algorithm" 
using namespace std;
const int MAXN = 25;
int arr[MAXN];
bool open[MAXN];
int main() {
    int n, m, k, q, ca = 1;
    while (scanf("%d%d%d", &n, &m, &k) && (n || m || k)) {
        memset(open, false, sizeof(open));
        int sum = 0, mx = 0;
        for (int i = 1; i <= n; i++)
            scanf("%d", &arr[i]);
        for (int i = 1; i <= m; i++) {
            scanf("%d", &q);
            if (open[q]) {
                sum -= arr[q];
            } else {
                sum += arr[q];
            }
            open[q] = !open[q];
            mx = max(mx, sum);
        }
        printf("Sequence %d\n", ca++);
        if (mx <= k) {
            printf("Fuse was not blown.\n");
            printf("Maximal power consumption was %d amperes.\n", mx);
        } else {
            printf("Fuse was blown.\n");
        }
        puts("");
    }
    return 0;
}
    View Code

     

J .Problem J(ZOJ1195普通签到题)

sol：写一个翻转数字的自定义函数，剩下的就好办了。全写主函数里会有点复杂

  • 水题 
    #include "cstdio"
#include "cstring"
#include "algorithm" 
using namespace std;
int rev(int k) {
    int res = 0;
    while (k) {
        res = res * 10 + k % 10;
        k /= 10;
    }
    return res;
}
int main() {
    int t, a, b;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &a, &b);
        printf("%d\n", rev(rev(a) + rev(b)));
    }
    return 0;
}
    View Code

     

I .Problem I(POJ1877贪心)

sol：sort一下，然后每次都往海拔最低的地方注水。特判k == 0的情况。

ps：让我好好吐槽一下。这题真心坑，建议直接去POJ提交，而且要选择C++编译器不要G++编译器。在HUD上提交一次AC之后再次提交一样的代码就wa了。说好的"The final value in each region description is an integer"，integer就这么变成了double。莫名其妙还浪费时间

  • 贪心 
    #include "cstdio"
#include "algorithm"
using namespace std;
const int MAXN = 905;
const int INF = 0x3f3f3f3f;
double arr[MAXN], k;
int main() {
    int n, m, ca = 1;
    while (scanf("%d%d", &n, &m) && (n || m)) {
        for (int i = 0; i < n * m; i++) 
            scanf("%lf", &arr[i]);
        sort(arr, arr + n * m);
        // 防越界 
        arr[n * m] = INF;
        scanf("%lf", &k);
        k /= 100;
        printf("Region %d\n", ca++);
        // 特判 
        if (k == 0) {
            printf("Water level is %.2lf meters.\n", arr[0]);
            printf("0.00 percent of the region is under water.\n");
            continue;
        }
        for (int i = 1; i <= n * m; i++) {
            if (i * (arr[i] - arr[i - 1]) >= k) {
                printf("Water level is %.2lf meters.\n", arr[i - 1] + k / i);
                printf("%.2lf percent of the region is under water.\n", i * 100.0 / n / m);
                break;
            } else {
                k -= i * (arr[i] - arr[i - 1]);
            }
        } 
    }
    return 0;
}
    View Code

     

K .Problem K(ZOJ1331暴力求解、计数排序)

sol：由于数组中的数最多99，可以用计数排序。

  • 计数排序 
    #include "cstdio"
#include "cstring"
#include "algorithm" 
using namespace std;
const int MAXN = 105;
int cnt[MAXN];
int main() {
    int n;
    while (scanf("%d", &n) && ~n) {
        if (n == 0) {
            int sum = 0;
            for (int i = 1; i < 50; i++) {
                if (cnt[i << 1]) {
                    sum += cnt[i];
                }
            }
            printf("%d\n", sum);
            memset(cnt, 0, sizeof(cnt));
        } else {
            cnt[n]++;
        }
    }
    return 0;
}
    View Code

     

posted @ 2019-04-25 20:19  Jathon-cnblogs  阅读(245)  评论(0)  编辑  收藏  举报