其他

Smallest DifferenceTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2679 Accepted: 766DescriptionGiven a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some

HopscotchTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1462 Accepted: 1043DescriptionThe cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows create a 5x5

/*The describe is not clear*/#include<cstdio>#include<cstring>using namespace std;const int maxn = 205;int degree[maxn];int f[maxn];int N,R;int find(int x){ if(x != f[x]) { f[x] = find(f[x]); } return f[x];}int main(){

B. Ilya and Queriestime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputIlya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.You'

    這道題剛開始糾結於輸入，還是題刷少了。囧！本來想用strchr函數和指標來操作的，但是在輸入的時候很糾結，就放棄了。這是poj 2255的加強版，不用真正的建樹後在遍曆，直接類比建樹的過程就解決了。    下面是代碼：  #include<stdio.h>#include<string.h>#define MAXN 10010int ans,flag,n;int s1[MAXN],s2[MAXN];int find_root(int b1,int e1,int

Shuffle'm UpTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4557 Accepted: 2167DescriptionA common pastime for poker players at a poker table is to shuffle stacks of chips. Shuffling chips is performed by starting with two stacks of poker

題目大意：就是先建兩個樹，然後遍曆這兩個樹。我就叫這種樹為四叉樹吧。難住我的地方就是遞迴建樹，然後是遍曆的時候想多來。具體的看代碼的注釋吧。下面是代碼：#include<stdio.h>#include<string.h>#include<iostream>#define MAXN 100000using namespace std;char s1[MAXN],s2[MAXN];char *s;typedef struct node{ node

Window PainsTime Limit: 2 Seconds      Memory Limit: 65536 KBBoudreaux likes to multitask, especially when it comes to using his computer. Never satisfied with just running one application at a time, he usually runs nine applications, each in its

PotsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8211 Accepted: 3481 Special JudgeDescriptionYou are given two pots, having the volume of A and B liters respectively. The following operations can be performed:FILL(i)        fill the pot 

分析：這道題蠻有意思的，雖然說很簡單，但是可以當做練手吧，鍛煉自己碼代碼的能力。注意：當a和b相等的時候，判定這個輸入不合法，不做任何的操作。#include<stdio.h>#include<string.h>#define MAXN 26int blocks[MAXN][MAXN];int n,order[MAXN],position[MAXN];void init(){ memset(blocks,-1,sizeof(blocks)); for(int

因為沒有權值，所以不用使用Floyd演算法求最短路。直接用bfs做。。。#include<cstdio>#include<cstring>using namespace std;const int maxn = 21;int G[maxn][maxn];int vis[maxn][maxn];struct node{ int level; int u;} q[maxn*maxn];int bfs(int s,int e){ int front = 1;

模板題，可以用kruskal，也可以用最短路。#include<cstdio>#include<string>#include<algorithm>#include<cmath>using namespace std;const int maxn = 5555;int u[maxn],v[maxn],p[maxn];double map[105][2];struct node{ int id; double w;}

/* 就是一些數字字串 ，怎樣排序讓組成的數字最大 貪心： 每次選兩個字串組成的數大的放在前面就行了 */ #include<stdio.h>#include<string.h>#include<iostream>using namespace std;//int str[55][1000];struct node{ char str[128]; // int len;}ss[57];bool cmp(node a,node b){

/*The analysis of this problem What is a Carmichael number? First it is not a prime; Second it must meet the contition that a^p mod p=a.*/#include<iostream>#include<cstdio>#include<cstdlib>//get a random

一、數論初步        1）、歐幾裡德演算法及其擴充                  歐幾裡德演算法也叫輾轉相除法。根據gcd(a,b)=gcd(b,a%b)可以由遞迴寫出該演算法。                  int gcd(int a,int b) { return b==0?a:gcd(b,a%b); }        2）、素數相關                   可用Eratosthenes篩法構造素數表。 void

                                                                      給定的m*n的棋盤，放置黑白皇后的方式從三個方面考慮。行，列，對角線。行：選取一個皇后放置一個皇后有m*n中方式，接下來放置另一個皇后，有(n-1)中方式，所以按行考慮有n*m*（n-1）種方式。列：同理可得有n*m*（m-1）中方式。對角線： 對角線有兩個方向的，”/“和”\“兩個方向，先考慮一個方向。                 每條對角線的長度為：

題目大意：從根結點開始標號，向左標號減一，向右標號加一。下面是代碼：#include<stdio.h>#include<string.h>#define MAXN 1000int ans[MAXN];int T;int build(int n,int num){ ans[n]+=num; int s; scanf("%d",&s); if(s!=-1) build(n-1,s); scanf("%d",&s);

題目很簡單，資料不大，直接深搜。#include<stdio.h>#include<string.h>#define MAXN 105int map[MAXN][MAXN];int n,m;void dfs(int i,int j){ if(!map[i][j]) return; map[i][j]=0; dfs(i+1,j); dfs(i-1,j); dfs(i-1,j-1); dfs(i-1,j+1); dfs(i+1,j-1

MatrixTime Limit: 6000MS Memory Limit: 65536KTotal Submissions: 4150 Accepted: 1007DescriptionGiven a N × N matrix A, whose element in the i-th row andj-th columnAij is an number that equals i2 + 100000 ×i +j2 - 100000 × j + i × j, you are to find

文章目錄 Sample InputPossible Output for Sample Input Problem E: WeddingUp to thirty couples will attend a wedding feast, at which they will be seated on either side of a long table. The bride and groom sit at one end,