[C++知识库]20210827阿里C/C++开发笔试

题目1.

给出一个数组a，需要对其做k次操作，每次操作是让其中的一个元素的值+1，k次操作之后使得数组的最大值最小。

#include <stdio.h>

int main()
{
    const int N = 100000 + 5;
    long long a[N];
    int n;
    long long k;
    long long maxn = 0;
    scanf("%d %lld", &n, &k);
    for (int i = 0; i < n; i++) {
        scanf("%lld", &a[i]);
        maxn = maxn > a[i] ? maxn : a[i];
    }
    long long sum = k;
    for (int i = 0; i < n; ++i) {
        long long x = maxn - a[i];
        sum -= x;
        if (sum <= 0) {
            break;
        }
    }
    if (sum > 0) {
        long long ans = (maxn + sum/n);
        long long rest = sum%n;
        if (rest > 0) ans++;
        printf("%lld\n", ans);
    }
    else {
        printf("%lld\n", maxn);
    }
    return 0;
}

题目2.

（笔试过程中并没有做出来，但是样例过了）

给出一个n*m的地图还给出k支部队，地图上1表示沼泽 不能走，0表示道路 能走，每支部队过1单位的时间能走1格，问这k支部队最快多久能汇合，如果不能汇合则输出-1。

#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string.h>
using namespace std;

struct node {
    int x, y;
    int cost;
};

int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
int vis[505][505] = {0};
int tmpVis[105][105];
int maps[505][505];

int main()
{
    int res = 0x3f3f3f3f;
    memset(tmpVis, 0x3f3f3f3f, sizeof(tmpVis));
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j)
            scanf("%d", &maps[i][j]);
    }
    int k;
    scanf("%d", &k);
    while (k--) {
        int x, y;
        scanf("%d%d", &x, &y);
        queue<node> q;
        node first = node{x, y, 0};
        q.push(first);
        memset(tmpVis, 0x3f3f3f3f, sizeof(tmpVis));
        tmpVis[x][y] = 0;
        while (!q.empty()) {
            node head = q.front();
            q.pop();
            for (int i = 0; i < 4; ++i) {
                int tmpx = head.x + dx[i];
                int tmpy = head.y + dy[i];
                if (tmpx <= 0 || tmpx > n || tmpy <= 0 || tmpy > m)
                    continue;
                int tmpCost = head.cost + 1;
                if (maps[tmpx][tmpy] == 0 && tmpCost < tmpVis[tmpx][tmpy]) {
                    tmpVis[tmpx][tmpy] = tmpCost;
                    node tmp = node{tmpx, tmpy, tmpCost};
                    q.push(tmp);
                }
            }
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                vis[i][j] = max(vis[i][j], tmpVis[i][j]);
            }
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            res = min(res, vis[i][j]);
        }
    }
    if (res == 0x3f3f3f3f) res = -1;
    printf("%d\n", res);
    return 0;
}