dmxqf44884-优快云博客

转载九度OJ 1004

#include<stdio.h> #include<algorithm> using namespace std; int a[2000005]; int main() { int n,m; while(scanf("%d",&n)!=EOF) { for(int i=0; i<n; i...

2014-04-15 23:24:00 127

转载九度OJ 1003

#include<stdio.h> #include<string> #include<iostream> using namespace std; int main() { char a[20],b[20]; while(scanf("%s%s",a,b) != EOF) { int na...

2014-04-12 23:54:00 133

转载九度OJ 1002

#include<stdio.h> #include<math.h> #include<algorithm> using namespace std; int main() { int p,t,g1,g2,g3,gj; while(scanf("%d %d %d %d %d %d",&p,&t,&am...

2014-04-11 13:35:00 125

转载九度OJ 1553

http://ac.jobdu.com/problem.php?pid=1553 /* 时针和分针位mod12，之和再mod6计算大角度，时针和分针的位置关系分4种情况讨论然后加上时针小角度的转动。 */ #include<stdio.h> int main() { int h,m; while(scanf("%d%*c...

2014-04-04 07:28:00 113

转载 codeforces 141A

1 #include<iostream> 2 #include<vector> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 string sortString(string s) 7 { 8 for(int i=0;...

2013-08-28 11:11:00 138

转载 codeforces 50A

http://codeforces.com/problemset/problem/50/A #include<cstdio> #include<iostream> using namespace std; int main() { int n,m,i; while(cin>>m>>n) { if(n%2==0) co...

2013-08-20 20:34:00 213

转载 codeforces 118A

http://codeforces.com/problemset/problem/118/A 1 #include<iostream> 2 #include<string> 3 #include<queue> 4 using namespace std; 5 string a; 6 queue<char>b; ...

2013-08-12 23:05:00 161

转载 codeforces 71A

http://codeforces.com/problemset/problem/71/A 1 #include<iostream> 2 #include<string> 3 using namespace std; 4 string a; 5 int main() 6 { 7 int n; 8 cin>&...

2013-08-12 22:30:00 151

转载 hdu1175 bfs

// hdu 1175 #include <iostream> #include <cstring> #include <cstdio> #include <queue> using namespace std; struct node { int x , y ; int step ; }s,e; ...

2013-07-26 16:10:00 94

转载 priority_queue

#include<iostream> #include<functional> #include<queue> using namespace std; struct node { friend bool operator<(node n1,node n2) { return n1.priori...

2013-07-25 23:21:00 75

转载最小生成树-Kruskal算法模板

#include<iostream> #include<cstdio> #include<algorithm> using namespace std; #define maxn 11 #define maxm 20 struct edge { int u,v,w; }edges[maxm]; int father[ma...

2013-07-22 23:18:00 88

转载 hdu 1213 并查集

http://acm.hdu.edu.cn/showproblem.php?pid=1213 #include<iostream> #include<cstdio> using namespace std; #define N 1000 int father[N]; void ufset() { for(int i=0;i<N;i...

2013-07-22 11:36:00 87

转载 hdu1702 list or stack+queue

http://acm.hdu.edu.cn/showproblem.php?pid=1702 #include<iostream> #include<cstdio> #include<cstring> #include<list> using namespace std; int main() { // freop...

2013-07-22 00:14:00 92

转载 codeforces 330B

http://codeforces.com/problemset/problem/330/B 题解: 关键在于 m < n/2，即必存在一个点能跟所有点相连,那么只用找出这么一个能连接所有点的点。然后输出的边肯定只有n-1条。 #include<iostream> #include<cstdio> #include<cstring&gt...

2013-07-21 10:26:00 165

转载 codeforces 330A

http://codeforces.com/problemset/problem/330/A #include<iostream> #include<cstdio> #include<cstring> using namespace std; int row[10][10],col[10][10]; char s[10][10]; i...

2013-07-21 09:39:00 145

转载 hdu1394 线段树求最小逆序数

hdu 1394 http://acm.hdu.edu.cn/showproblem.php?pid=1394 用线段树求逆序数，例如要求x的逆序数只需要访问（x+1，n）段有多少个数，就是x的逆序数。还有就是求最小逆序数的时候有个巧妙的想法，当把x放入数组的后面，此时的逆序数应该为x没放入最后面之前的逆序总数加上（n-x）再减去（x-1）；sum = sum+(n-x[i])-...

2013-07-21 01:58:00 181

转载 Hdu1754-线段树-单点更新

#include<iostream> #include<cstdio> #include<algorithm> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 const int maxn=200005; int MA...

2013-07-20 16:28:00 79

转载 Hdu1166-- 线段树模板

#include<iostream> #include<cstdio> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 const int maxn=555555; int sum[maxn<<2]; void pus...

2013-07-20 15:56:00 102

转载 Largest product in a grid（ Project Euler problem 11）

In the 2020 grid below, four numbers along a diagonal line have been marked in red. 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 0849 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 ...

2013-07-17 09:10:00 143

转载 Smallest multiple

#include<iostream> #include<algorithm> using namespace std; int lcm(int a ,int b) { int big=max(a,b); int small=min(a,b); int max=a*b; for(int i=big; i<ma...

2013-07-15 21:59:00 201

转载 Largest palindrome product

https://projecteuler.net/problem=4 1、它是回文数 2、存在2个三位数乘积等于它。 1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 bool is_ok(int x) 5 { 6 if(x%10!=x/10...

2013-07-15 20:37:00 62

转载 codeforces-320A-Magic Numbers

http://codeforces.com/problemset/problem/320/A A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 1414...

2013-06-27 23:37:00 210