做题笔记：某大讯飞真题28道

cpp 复制代码

 #include <stdio.h>
    int main() {
        int n;
        scanf("%d", &n);
        if (n % 2 != 0) {
            printf("-1\n");
            return 0;
        }
        int k = n/2;   // n为偶数
        int y_max = n/8;  // 8个袋子的最大数量
        int ans = -1;
        for (int y = y_max; y>=0; y--) {
            // 计算剩余的橘子数=2k-8y = 4k-8y? 注意：方程是3x+4y=k
            if ( (k - 4*y) >=0 && (k-4*y)%3 ==0 ) {
                int x = (k-4*y)/3;
                ans = x+y;
                break;
            }
        }
        printf("%d\n", ans);
        return 0;
    }

cpp 复制代码

#include <stdio.h>
#include <stdlib.h>
#define MAX_N 100010

int a[MAX_N], b[MAX_N];
int tree[4 * MAX_N];

// 手动实现二分查找（替代lower_bound）
int binary_search(int arr[], int n, int x) {
    int l = 0, r = n - 1;
    while (l <= r) {
        int mid = l + (r - l) / 2;
        if (arr[mid] < x) l = mid + 1;
        else if (arr[mid] > x) r = mid - 1;
        else return mid; // 找到精确匹配
    }
    return l; // 返回第一个不小于x的位置
}

// 线段树相关函数保持不变（同原代码）
void build(int node, int l, int r) {
    tree[node] = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(node << 1, l, mid);
    build(node << 1 | 1, mid + 1, r);
}

void update(int node, int l, int r, int idx, int delta) {
    if (l == r) {
        tree[node] += delta;
        return;
    }
    int mid = (l + r) >> 1;
    if (idx <= mid) update(node << 1, l, mid, idx, delta);
    else update(node << 1 | 1, mid + 1, r, idx, delta);
    tree[node] = tree[node << 1] + tree[node << 1 | 1];
}

int query(int node, int l, int r, int k) {
    if (l == r) return l;
    int mid = (l + r) >> 1;
    if (tree[node << 1] >= k)
        return query(node << 1, l, mid, k);
    else
        return query(node << 1 | 1, mid + 1, r, k - tree[node << 1]);
}


int cmp(const void* a, const void* b) {
    return *(int*)a - *(int*)b;
}

int main() {
    int n, m;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
        b[i] = a[i];
    }

    // 离散化处理
    qsort(b, n, sizeof(int), cmp);
    m = 1;
    for (int i = 1; i < n; i++)
        if (b[i] != b[i - 1]) b[m++] = b[i];

    // 构建权值线段树
    build(1, 0, m - 1);

    // 初始化更新（使用binary_search替代lower_bound）
    for (int i = 0; i < n; i++) {
        int pos = binary_search(b, m, a[i]);
        update(1, 0, m - 1, pos, 1);
    }

    int len = n;
    for (int i = 0; i < n; i++) {
        int k = (len + 1) / 2;  // 中位数位置
        int idx = query(1, 0, m - 1, k);
        printf("%d ", b[idx]);
        update(1, 0, m - 1, idx, -1);  // 删除当前中位数
        len--;
    }
    return 0;
}

cpp 复制代码

    #include <stdio.h>
    #include <stdlib.h>
    #define ll long long

    ll terms[100]; // will not exceed 30 items

    int main() {
        int T;
        scanf("%d",&T);
        while (T--) {
            ll n;
            scanf("%lld",&n);
            ll p = 1;
            int count = 0;
            while (n>0) {
                ll r = n % 3;
                if (r==0) {
                    n /= 3;
                } else if (r==1) {
                    terms[count++] = p;
                    n = (n-1)/3;
                } else if (r==2) {
                    terms[count++] = 2 * p;
                    n = (n-2)/3;
                }
                p *= 3;
            }
            // reverse terms array: index 0 to count-1 (originally added in increasing b, so we output from count-1 to 0)
            // but the problem requires from large to small.
            printf("%d\n",count);
            if (count==0) {
                // no terms, but count=0, so just output the empty line?
                printf("\n");
            } else {
                // output from the last term to the first term
                for (int i=count-1; i>=0; i--) {
                    printf("%lld ", terms[i]);
                }
                printf("\n");
            }
        }
        return 0;
    }