做题笔记：某大讯飞真题28道

 #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=kif ( (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;}

#include <stdio.h>
#include <stdlib.h>
#define MAX_N 100010int 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);elsereturn 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;
}

    #include <stdio.h>#include <stdlib.h>#define ll long longll terms[100]; // will not exceed 30 itemsint 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 termfor (int i=count-1; i>=0; i--) {printf("%lld ", terms[i]);}printf("\n");}}return 0;}