A place to record sth.

题意：手机图案解锁，给定n个数字，求可以得到的图案有多少种。其中有一些数字需要借助别的数字才能到达。#include#include#include#includeusing namespace std;int vis[300],a[11],mp[11][11];int ans[400000][11],f[11];int n,len;void init(){ mp[1][

题意：找出不同的数字。#include#include #includeusing namespace std;int main(){ int T,n,i,a[105]; scanf("%d",&T); while(T--) { scanf("%d",&n); for(i=0; i<n; i++) {

题意：宽度从3到10输出N形状的图形。#include#includeusing namespace std;char str[16][16];int t;void display(int n){ int i,j; for(i=0;i<16;++i){ for(j=0;j<16;++j){ str[i][j]=' ';

题意：计算每个座位的value值，根据坐在该座位的队伍长度与其上下左右方向的四个队伍长度比较。若该位置的队伍长度比一个方向的长度大，则权值减去长度差的绝对值；反之则加长度差的绝对值。#include#include#includeusing namespace std;int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};int main(){

题意：字符串转换后倒序输出。#include #include #include using namespace std; char arr[310]; int main() { char s[105]; int i,j,t; arr[ int('n') ]='u'; arr[ int('5') ]='s';

Mine NumberTime Limit: 1000MS Memory limit: 65536K题目描述Every one once played the game called Mine Sweeping, here I change the rule. You are given an n*m map, every element is a '*' repres

Cylinder CandyTime Limit: 2 Seconds      Memory Limit: 65536 KB      Special JudgeEdward the confectioner is making a new batch of chocolate covered candy. Each candy center is shaped as a c

http://acm.hdu.edu.cn/showproblem.php?pid=4715#include#include#include #include #include #include using namespace std;#define N 1000000bool isprm[N];long long xx[N];void isprime(){ i

http://acm.hdu.edu.cn/showproblem.php?pid=4709思路：海伦公式不可以，精度不够。#include#include#includeusing namespace std;double X[128];double Y[128];double ff(int i,int j,int k){ return fabs(X[i]*Y[

http://acm.hdu.edu.cn/showproblem.php?pid=4707#include#include#include#includeusing namespace std;#define MAXN 200005int sum;int D;int dis[MAXN];bool vis[MAXN];struct Edge{ int to

http://acm.hdu.edu.cn/showproblem.php?pid=4706#include#includeusing namespace std;char str[16][16];int t;void display(int n){ int i,j; for(i=0;i<16;++i){ for(j=0;j<16;++j){

Rotation Lock PuzzleTime Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u DescriptionAlice was felling into a cave. She found a strange door with a number square

Pick applesTime Limit: 1000MS Memory limit: 165536K题目描述Once ago, there is a mystery yard which only produces three kinds of apples. The number of each kind is infinite. A girl carrying a

Pixel densityTime Limit: 1000MS Memory limit: 65536K题目描述 Pixels per inch (PPI) or pixel density is a measurement of the resolution of devices in various contexts; typically computer di

Fruit Ninja IITime Limit: 5000ms   Memory limit: 65536K  有疑问？点这里^_^题目描述 Have you ever played a popular game named "Fruit Ninja"?Fruit Ninja (known as Fruit Ninja HD on the iPad

琪露诺的算术教室Time Limit: 1000msMemory Limit: 65536KBDescription给出一个非负整数A，将这个数字的最低位移动到最高位（原来的最高位变为次高位，次低位变成最低位），得到非负整数B，发现B恰好是A的k倍。现给出A的最低位的值n，和倍数k，求最小的非负整数B。Input第一行输入一个正整数T

题意：一个三角形的结构，第一行有一个，第二行有二个……，你处于第一行，需要达到最后一行的第一个，最左边的只能向左下和右下走，其他的可以向左、向左下和向右下走，求达到目的地的长度的期望。思路：对于数学期望的定义是这样的。数学期望E(X) = X1*p(X1) + X2*p(X2) + …… + Xn*p(Xn)比如打靶打中8环的概率为0.3  ,打中7环的概率为0.7，那

题意：给长度为N的字符串，只存在小写字母。问你山形的子序列有多少种，这里是子序列不是子串。所谓的山形就是严格的递增到递减。思路：从左到右求每一个位置递增子序列的个数，然后再从右向左求每一个位置递增子序列的个数，然后对应位置相乘就可以得到最终结果。#include #include #include #include using namespace std;int n;const

题意： 有一个式子(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1)，给定一个P，求x^P的系数。思路：根据推导得出：例如 x^3     3的二进制表示为11，则x^3的系数为a[0]*a[1]        x^4     4的二进制表示为100，则x^4的系数为a[2]对于本题输入n个a

题意：TeamA request B pages。按照这个形式输入第A个队伍需要打印B张纸。然后定义s=(s*x+y)%mod。当打印的纸张数>=s时，便会重新打印这个队伍的纸张。按照要求输出。#include #include using namespace std;struct p{ int num; char name[25];} a[1005];int

题意：已知一个等边三角形的两个顶点a(x1,y1)和b(x2,y2)，求第三个顶点，按照逆时针顺序求。思路：向量的旋转。参考博客：http://www.cnblogs.com/E-star/archive/2013/06/11/3131563.html#include #include #include using namespace std;int main(){

题意：n，m个男女，凑成n对couple，最后分成11个组，检测能否分成。思路：java大数。n==m才可以成couple，最后mod 11。import java.math.BigInteger;import java.util.Scanner;public class Main {//类名要用Main public static void main(String[] ar

思路：当石子个数大于2时，后手赢。#include#includeusing namespace std;int main(){ int T; long long N; scanf("%d",&T); while(T--){ scanf("%lld",&N); if(N>2){ print

题意：查找只被竞价一次的价格中的最小值。set用法:http://www.cnblogs.com/BeyondAnyTime/archive/2012/08/13/2636375.html#include #include #include #include #include using namespace std;int main(){ set myset;

题意：n个单位，每个单位每秒增加1法力，在某些时间取走一些区间的法力值（取走之后该区间所有单位的法力变为0），求取得的所有法力值。思路：线段树区间更新，求和。区间求和之后要把该区间清零。set+add操作，set在先，add在后。#include#include#includeusing namespace std;#define L(root) ((root)<<1)#def

题意：n个数围成一个圈，编号0~n-1，从一个数到其上一个和下一个的数的概率相同（即都是0.5）。给出n，求从0出发到达一个数x所需要的步数的数学期望。思路：这里用递推可以水一下。首先可以确定三个结论d[i]=0.5*d[i+1]+0.5*d[i-1]d[n]=d[0]=0d[i]=d[n-i]规律：d[0]=0,d[1]=n-1,d[2]=(n-1)+(n-3),d[

Hearthstone II Time Limit: 2000MS Memory limit: 65536K 题目描述The new season has begun, you have n competitions and m well prepared decks during the new season. Each competition you could

Full Binary TreeTime Limit: 2000MS Memory limit: 65536K题目描述In computer science, a binary tree is a tree data structure in which each node has at most two children. Consider an infinite

FactorialTime Limit: 2000ms   Memory limit: 65536K  有疑问？点这里^_^题目描述Homelesser hates mathematics. He cannot even do addition and subtraction, not to mention computing factorial. He asks

angry_birds_again_and_againTime Limit: 2000MS Memory limit: 65536K题目描述The problems called "Angry Birds" and "Angry Birds Again and Again" has been solved by many teams in the series of

Weighted MedianTime Limit: 2000ms   Memory limit: 65536K  有疑问？点这里^_^题目描述For n elements x1, x2, ..., xn with positive integer weights w1, w2, ..., wn. The weighted median is the element x

http://www.cnblogs.com/en-heng/p/4002658.html

http://blog.youkuaiyun.com/dark_scope/article/details/8880547

http://acm.sdut.edu.cn/sdutoj/contest_show.php?contest_id=1254http://acm.sdut.edu.cn/sdutoj/contest_show.php?contest_id=1255http://acm.sdut.edu.cn/sdutoj/contest_show.php?contest_id=1256http://a