Duke10的博客

python3做法：n = int(input())lst= list(map(int,input().split()))times = []for i in lst: for j in lst: if i < j and i + j in lst and i+j not in times: times.append(i+j)print(len

#include<stdio.h>struct Node // 定义结构体，方便处理数据{ int start; //定义开始时间 int end; //定义结束时间}a[200];void quick_sort(struct Node *arr, int l, int r) //对结束时间进行排序{ if (l < r) //进行判断，用来终止递

#include<stdio.h>int sum;void f(int *a, int w, int n){ int i = 0, j = 1, temp = 0, t, temp1, s; for (i = 0; i < n; i++) { s = 0, temp1 = 0; for (j = i + 1; j < n; j++)

#include<stdio.h>#include<string.h>int main(){ char a[5000], b[5000], temp; int len, i, t; while (scanf("%s%s", a, b) != EOF) { int c = 0, d = 0, num = 0; len = strle

#include<stdio.h>int main(){ int a,i; while (scanf("%d",&a) != EOF) { int b[100]={0}; if (a == 0) { printf("%d\n",a); continue; }

#include<stdio.h>#include<string.h>int main(){ int i,j,n,a,b,c,t=0; char str[100]={0}; scanf("%d",&n); n=n+1; while (n--) { gets(str); a=strlen(str);

#include<stdio.h>int main(){ int a[100][100]={0}; int k,m,n,i,j; scanf("%d",&n); m=n/2-1; i=0; j=n-1; for (k=1;k<=n*n;k++) { a[i][j]=k; if (j==n-1 && i

无线网络覆盖 时间限制：3000 ms | 内存限制：65535 KB 难度：3描述我们的乐乐同学对于网络可算得上是情有独钟，他有一个计划，那就是用无线网覆盖郑州大学。现在学校给了他一个机会，因此他要购买很多的无线路由。现在他正在部署某条大道的网络，而学校只允许把他的无线路由器放在路的正中间。我们默认这条大道是笔直的并且它在任何地方的宽度都一样。并且所有的路由器的覆盖面积是相同的。现在乐乐计

#include<stdio.h>#include<string.h>#define MAX(a,b) (a > b ? a : b)int main(){ int n, v; while (~scanf("%d%d",&n, &v) && n + v != 0) { int i, j, volume, value, dp[v+1];

#include&lt;stdio.h&gt;int main(){ int num; while (num != 0) { scanf("%d", &amp;num); int a[100], i, temp; for (i = 0; i &lt; num; i++) { sc...

街区最短路径问题时间限制：3000 ms | 内存限制：65535 KB难度：4描述一个街区有很多住户，街区的街道只能为东西、南北两种方向。住户只可以沿着街道行走。各个街道之间的间隔相等。用(x,y)来表示住户坐在的街区。例如（4,20），表示用户在东西方向第4个街道，南北方向第20个街道。现在要建一个邮局，使得各个住户到邮局的距离之和最少。求现在这个邮局应该建...

A + B Problem IITime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 407062 Accepted Submission(s): 78858Problem Description I have a very ...

Max SumTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 277191 Accepted Submission(s): 65811Problem Description Given a sequence a[1],a...

Let the Balloon RiseTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 137506 Accepted Submission(s): 54296Problem Description Contest ti...

回文数猜想Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8092 Accepted Submission(s): 4955Problem Description 一个正整数，如果从左向右读（称之为正序数）和从右向左读（...

#include<stdio.h>int dp[25][25];int maxx(int a,int b){ if(a>b) return a; else return b;}int main(){ int n,m; while(scanf("%d%d",&n,&m)!=EOF) { int i,j;

#include<stdio.h>#include<math.h>void quick_sort(double s[], int l, int r){ if (l < r) { double x = s[l]; int i = l, j = r; while (i < j) { while(

#include<stdio.h>typedef struct { char name[20]; //姓名 int grade1; //平均成绩 int grade2; //班级评议成绩 char p; //是否为学生干部 char q; //是否为西部学生 int num; //论文数量 i

#include <stdio.h> #include <string.h> #include <stack> using namespace std; stack <char> Q; char s[11000]; int main() { int T,i; Q.push('#'); scanf("%d",&T); whil

#include<stdio.h>#include<string.h>int dp[1010],a[1010];int main(){ int t,n; scanf("%d",&t); while(t--) { scanf("%d",&n); memset(dp,0,sizeof(dp)); int max,ma

#include<stdio.h>#include<string.h>#define Min(a,b) (a < b ? a : b)int main(){ int n, t; while (~scanf("%d%d",&n,&t) && n + t != 0) { int i, j, a[100] = {0}; for (i = 1;

#include<stdio.h>#include<string.h>#include<stdlib.h>#define MAX(a,b) a > b ? a : bint main(){ int n; scanf("%d",&n); while (n--) { char a[1000] = {0}, b[1000] = {0};

#include<stdio.h>int main(){ int n; scanf("%d",&n); while (n--) { int len; scanf("%d",&len); int a[41], i; a[2] = 3; a[3] = 5; for (i =

#include<stdio.h>#include<stdlib.h>int n, m, i, j;int sort(const void *a, const void *b){ return *(int *)a - *(int *)b;}int judge(int *a, int mid){ int temp = a[0], count = 1; for (i

#include<stdio.h>#include<string.h>int main(){ char a[1001], n; char b[10][100] ={"bowl", "knife", "fork", "chopsticks"}; while (scanf("%d",&n) != EOF) { int i, j; fo

#encoding:utf-8lst = []lst1 = []s = 0while True: lst1 = list(input()) for i in lst1: if i != 'E': lst.append(i) else: s = 1 break if s

#include<stdio.h>int main(){ int i, n, m; int a[10] = {1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3991680}; scanf("%d",&n); while (n--) { scanf("%d",&m); for (i =

#include<stdio.h>#include<stdlib.h>struct Node{ double value; double weight;}a[10];void quick_sort(struct Node *arr, int l, int r) //对结束时间进行排序{ if (l < r) //进行判断，用来终止递归 {

#include<stdio.h>#include<stdlib.h>int sort(const void *a, const void *b){ return *(int *)a - *(int *)b;}int min(int a, int b){ if (a < b) return a; else return b;}i

#include<stdio.h>struct Node{ int start; int end;}a[10000];void quick_sort(struct Node *arr, int l, int r) //对结束时间进行排序{ if (l < r) //进行判断，用来终止递归 { int i = l, j = r, x

#include<stdio.h>#include<math.h>struct Node{ double start; // 横坐标 double end; //半径}a[10000];void quick_sort(struct Node *arr, int l, int r) //快速排序{ if (l < r)

士兵队列训练问题Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11213 Accepted Submission(s): 4955Problem Description 某部队进行新兵队列训练，将新兵从一开始按顺序依次...