题意：给你正视和側视图，求最多多少个，最少多少个
思路：贪心的思想。求最少的时候：由于能够想象着移动，尽量让两个视图的重叠。所以我们统计每一个视图不同高度的个数。然后计算。至于的话。就是每次拿正视图的高度去匹配側视求最大
#include#include #include #include using namespace std;const int MAXN = 1000;int k;int view[2][MAXN];int main() { int t; scanf("%d", &t); while (t--) { scanf("%d", &k); memset(view, 0, sizeof(view)); for (int i = 0; i < 2; i++) for (int j = 0; j < k; j++) { int x; scanf("%d", &x); view[i][x]++; } int Min = 0, Max = 0; for (int i = 1; i < MAXN; i++) Min += i * max(view[0][i], view[1][i]); for (int i = 1; i < MAXN; i++) for (int j = 1; j < MAXN; j++) Max += min(i, j)*view[0][i]*view[1][j]; printf("Matty needs at least %d blocks, and can add at most %d extra blocks.\n", Min, Max-Min); } return 0;}