qwerty559的博客

问题描述：有n个矩形，每个矩形可以用两个整数a、b描述，表示它的长和宽。矩形X(a,b)可以嵌套在矩形X(c,d)中，当且仅当a<c，b<d，或者a<d，b<c（相当于把矩形X旋转90°）。例如，(1,5)可以嵌套在(6,2)内，但不能嵌套在(3,4)内。你的任务是选出尽量多的矩形排成一行，使得除了最后一个之外，每一个矩形都可以嵌套在下一个矩形内。如果有多解，矩形编号...

#include<stdio.h>#include<string.h>int main(){ int T, n; int a[10]; scanf("%d", &T); while (T--) { memset(a, 0, sizeof(a)); scanf("%d", &n); for (int i = 1; i <= ...

#include<stdio.h>#include<string.h>#include<ctype.h>#define maxn 100int main(){ int T, a[4]; char s[maxn]; scanf("%d", &T); while (T--) { memset(a, 0, sizeof(a)); ...

#include<stdio.h>#include<string.h>#define maxn 100int main(){ int T; char s[maxn]; scanf("%d", &T); while (T--) { scanf("%s", s); int len = strlen(s); int ans = 0, w =...

#include<stdio.h>int main(){ int x, y, kase = 0, first = 1; char s[10][10], c; while (1) { c = getchar(); if (c == 'Z') break; int j = 0; do { if (c == ' ') { x = 0; ...

#include<stdio.h>#include<string.h>#include<math.h>const double eps = 1e-4;int main() { double M[20][40]; long long E[20][40]; // 打表 for (int i = 0; i <= 9; ++i) { f...

#include<stdio.h>#include<string.h>#define maxn 3005int main(){ int a, b; int ys[maxn], xs[maxn]; while (scanf("%d%d", &a, &b) == 2) { memset(ys, -1, sizeof(ys)); mems...

#include<stdio.h>#include<string.h>#define maxn 105int match(char s1[], char s2[]){ int len1 = strlen(s1); int len2 = strlen(s2); int k = -1; for (int i = 0; i < len1; ++i) {...

#include<stdio.h>int main(){ int w, h, t; int rect[6][2]; while (scanf("%d%d", &w, &h) == 2) { if (w < h) { t = w; w = h; h = t; } rect[0][0] = w; rect[0][1] = h; ...

#include<stdio.h>#include<string.h>#define maxn 100005int main(){ char s[maxn], t[maxn]; while (scanf("%s%s", s, t) == 2) { int len1 = strlen(s); int len2 = strlen(t); int i ...

#include<stdio.h>int main(){ int T, m, n; char s[55][1005], t[] = "ACGT"; scanf("%d", &T); while (T--) { scanf("%d%d", &m, &n); for (int i = 0; i < m; ++i) { scanf(...

#include<stdio.h>#include<string.h>int main(){ int r, c, kase = 0, first = 1; int a[15][15]; char s[15][15]; while (scanf("%d", &r) == 1 && r) { int cnt = 0; memse...

用1,2,3,...,9组成三位数abc，def，和ghi，每个数字恰好使用一次，要求abc:def:ghi=1:2:3。按照“abc def ghi”的格式输出所有解，每行一个解。#include<stdio.h>int main(){ int a[3]; for (int i = 123; i <= 321; ++i) { int b[10] = {0};...

输入正整数a，b，c，输出a/b的小数形式，精确到小数点后c位。，。输入包含多组数据，结束标记为a=b=c=0。样例输入：1 6 40 0 0样例输出：Case 1: 0.1667#include<stdio.h>int main(){ int a, b, c, kase = 0; while (scanf("%d%d%d", &a, &a...

#include<stdio.h>#include<string.h>#define maxn 100int main(){ int T, first = 1; char s[maxn]; scanf("%d", &T); while (T--) { scanf("%s", s); int len = strlen(s); int f...