Codeforces Round #731 (Div. 3) A - D 题解

A.Shortest Path with Obstacle

思路

起点、终点和障碍在同一直线，且障碍在起点终点之间则要多走 $2$ 步绕开障碍

Accepted code

#include <bits/stdc++.h>
using namespace std;

int t;
int sx, sy, fx, fy, tx, ty;

int main() {
    
    
    scanf("%d", &t);
    while (t--) {
    
    
        scanf("%d%d", &sx, &sy);
        scanf("%d%d", &tx, &ty);
        scanf("%d%d", &fx, &fy);
        int ans = abs(sx - tx) + abs(sy - ty);
        if (sx == fx && sx == tx && fy > min(sy, ty) && fy < max(sy, ty))
            ans += 2;

        if (sy == fy && sy == ty && fx < max(sx, tx) && fx > min(sx, tx))
            ans += 2;

        printf("%d\n", ans);
    }
    return 0;
}

B.Alphabetical Strings

思路

我是从 a 的两侧开始找，用两个指针找下一个字母，注意指针越界的情况。

Accepted code

#include <bits/stdc++.h>
using namespace std;

int t;
int l, r;
int ok, flag, idx;
char now;
char s[35];

int main() {
    
    
    scanf("%d", &t);
    while (t--) {
    
    
        ok = 0;
        scanf("%s", s);
        int len = strlen(s);
        for (int i = 0; i < len; i++) {
    
    
            if (s[i] == 'a') {
    
    
                ok = 1;
                flag = i;
            }
        }
        now = 'a';
        idx = 0;
        if (ok) {
    
    
            l = flag - 1;
            r = flag + 1;

            while (1) {
    
    
                if (idx >= len - 1) {
    
    
                    break;
                }
                if (l >= 0 && s[l] == (char)(now + 1)) {
    
    
                    --l;
                    ++now;
                    ++idx;
                } else if (r < len && s[r] == (char)(now + 1)) {
    
    
                    ++r;
                    ++now;
                    ++idx;
                } else {
    
    
                    ok = 0;
                    break;
                }
            }

            if (ok)
                puts("YES");
            else
                puts("NO");
        } else {
    
    
            puts("NO");
        }
    }
    return 0;
}

C. Pair Programming

思路

数组 $a,b$ 各一个指针，优先选择数值小的，注意指针越界的情况。

Accepted code

#include <bits/stdc++.h>
using namespace std;

int t;
int k, n, m;
int a[105];
int b[105];
int ans[205];

int main() {
    
    
    scanf("%d", &t);
    while (t--) {
    
    
        scanf("%d%d%d", &k, &n, &m);
        for (int i = 0; i < n; i++) {
    
    
            scanf("%d", &a[i]);
        }
        for (int i = 0; i < m; i++) {
    
    
            scanf("%d", &b[i]);
        }
        int aidx = 0, bidx = 0, idx = 0;
        int ok = 1;
        while (1) {
    
    
            if (idx >= n + m) break;
            if (aidx < n && (bidx >= m || a[aidx] < b[bidx])) {
    
    
                if (a[aidx] == 0)
                    ++k;
                else if (a[aidx] > k) {
    
    
                    ok = 0;
                    break;
                }
                ans[idx] = a[aidx];
                ++aidx;
            } else if (bidx < m) {
    
    
                if (b[bidx] == 0)
                    ++k;
                else if (b[bidx] > k) {
    
    
                    ok = 0;
                    break;
                }
                ans[idx] = b[bidx];
                ++bidx;
            } else {
    
    
                ok = 0;
                break;
            }

            ++idx;
        }
        if (ok) {
    
    
            for (int i = 0; i < n + m; i++) {
    
    
                printf("%d%c", ans[i], i == n + m - 1 ? '\n' : ' ');
            }
        } else {
    
    
            puts("-1");
        }
    }
    return 0;
}

D. Co-growing Sequence

思路

对于每个数，都将其与前一个数相或求出当前数应当化为的数，再将求出的数与当前数异或，所得的数即为 $y_i$ 。

Accepted code

#include <bits/stdc++.h>
using namespace std;

int t;
int n;
int a[200005];
int ans[200005];

int main() {
    
    
    scanf("%d", &t);
    while (t--) {
    
    
        scanf("%d", &n);
        for (int i = 0; i < n; i++) {
    
    
            scanf("%d", &a[i]);
            if (i) {
    
    
                int to = (a[i] | a[i - 1]);
                ans[i] = a[i] ^ to;
                a[i] = to;
            }
        }
        for (int i = 0; i < n; i++) {
    
    
            printf("%d%c", ans[i], i == n - 1 ? '\n' : ' ');
        }
    }
    return 0;
}