其他

  簡單的計數，只要做到不重，不漏，就應該能過了。  下面是代碼，因為是厚白數上的例題，所以自己很沒譜。   #include<iostream>using namespace std;long long ans[10000010];int main(){ ans[3]=0; for(long long i=4;i<=10000010;i++) ans[i]=ans[i-1]+((i-1)*(i-2)/2-(i-1)/2)/2; int n;

水題一道，直接DFS。下面是代碼:#include<stdio.h>#include<string.h>char map[35][85];//int vis[35][85];int move[4][2]={{1,0},{-1,0},{0,1},{0,-1}};int h;void dfs(int x,int y){ for(int i=0;i<4;i++) { int dx=x+move[i][0]; int

刷這道題的時候發現自己和大牛的差距了。剛開始我想的是如何類比這個迷宮，想來想去沒想出什麼就去網上搜大牛的想法。結果是用方格來替'\'和'/'，和我想的剛好相反。可是我卻放棄了思考。哎，下次堅決不能這樣了。言歸正傳，剛開始時用一個2*2的方格來類比斜杠，但是要判斷遞迴的方向，很是麻煩。共有8個方向需要判斷。所以換了用3*3的來類比這個斜杠。這樣只需要記下遍曆過的方格的個數就可以得到答案了。下面是代碼：#include<stdio.h>#include<string.h>#d

Popular CowsTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 19089 Accepted: 7678DescriptionEvery cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <=

Problem C: Arctic NetworkThe Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio

  It's not a Bug, it's a Feature! It is a curious fact that consumers buying a new software product generally do not expect the software to be bug-free. Can you imagine buying a car whose steering wheel only turns to the right? Or a CD-player that

Problem DThe Tourist GuideInput: standard inputOutput: standard output Mr. G. works as a tourist guide. His current assignment is to take some tourists from one city to another. Some two-way roads connect the cities. For each pair of neighboring

這道題搜尋裡面套搜尋，注意如何判斷兩個X是否相連調試注意各種細節。代碼見下：#include<stdio.h>#include<string.h>#include<algorithm>#define MAXN 105using namespace std;char map[MAXN][MAXN];int vis[MAXN][MAXN];int visx[MAXN][MAXN];int ans[MAXN];int w,h;int cnt;int move[4][

尼瑪，原來國際象棋的騎士和我們象棋的馬的走法差不多啊。這道題直接bfs做。下面是代碼：#include<cstdio>#include<cstring>#include<iostream>#define MAXN 10using namespace std;struct node{ int x; int y; int step; node(){} node(int a,int b,int c):x(a),y(b),step(c){

先bs下自己吧。這道題可以先打表，在提交。剛開始寫的二分有點問題，我先找到的是n，然後再找m。其實是先找到m，然後再找n。另一種解法是解方程，解佩爾方程。不管是暴力還是解方程都要先找出運算式。由題意可得n*(n+1)/2=x*(x+1)/2-n*(n+1)+n；化簡可得n*n/2=x*x+x。然後將其變形為標準的佩爾方程，然後再用遞推求解。下面是二分的代碼：#include<iostream>#include<cstdio>using namespace

題目給的是個三維的圖形，那麼就開一個三維的數組，然後用bfs就ok啦！下面是代碼：#include<iostream>#include<stdio.h>#include<cstring>using namespace std;#define MAXN 31struct node{ int l; int r; int c; int t; node(){} node(int x,int y,int z,int

   這道題有點水，直接上代碼吧！  #include<stdio.h>#include<string.h>#define MAXN 3700int hash[MAXN];int main(){ int n,m,d,h; scanf("%d",&n); while(n--) { int ans=0; scanf("%d",&d); scanf("%d",&m);

/* 小小的吐槽一下，資料有點問題，題目說的以n=0，m=0 結束。可是測試的資料應該是n=0，m！=0結束的，Orz， wa了多次*/#include<cstdio>#include<cstring>using namespace std;const int maxn = 105;int G[maxn][maxn];int vis[maxn];int topo[maxn];int n,m,t;void dfs(int u){ vis[u] = 1;

/*連續兩道題目的資料有問題，Ora。剛開死將這道題想簡單了，直接類比做的，額...*/#include<cstdio>#include<cstring>#include<string>#include<iostream>#include<cstdlib>using namespace std;const int maxn = 1001;const int Min = -100000000;int

/*簡單的枚舉*/#include<cstdio>#include<cstring>using namespace std;int a[4][2];int main(){ int N; while(scanf("%d",&N) && N) { for(int i = 0; i < 2*N; i++) { int x,y;

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;}

  Problem B: Audiophobia Consider yourself lucky! Consider yourself lucky to be still breathing and having fun participating in this contest. But we apprehend that many of your descendants may not have this luxury. For, as you know, we are the

Problem EConnect the CampusInput: standard inputOutput: standard outputTime Limit: 2 secondsMany new buildings are under construction on the campus of the University of Waterloo. The university has hired bricklayers, electricians, plumbers, and a

                                                                      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：