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//存储结构struct PolyNode{ int coef;//系数 int expon;//指数 PolyNode * next;//指针域}//多项式相乘PolyNode* Multiply(PolyNode* L1,PolyNode* L2){ PolyNode *t1,*t2,*t,*rear,*L; int c,e; L=(...

两个大整数用单链表H1和H2表示，链表中每个结点存储一位数字（假设大整数存储是由低位到高位逆序存储的）。大整数相加代码如下：LNode * Add(LNode* H1,LNode* H2){ LNode *p1=H1->next,*p2= H2->next,*p=NULL; LNode *H3=(LNode*)malloc(sizeof(LNode));//创...

排序#include #include#includeusing namespace std;bool comp(string a,string b);int main(){ int n,m; string str[105]; while(cin>>n>>m){ for(int i=0;i<m;i++) cin>>st

#includeusing namespace std;int main(){ int n,m,a[105]; while(cin>>n>>m){ int i,j,t; for(i=0;i<n;i++) cin>>a[i]; if(n/2>m){ for(i=0;i<n-m;i++

题目：令Pi表示第i个素数。现任给两个正整数M 4，请输出PM到PN的所有素数。在第四个测试点跪了好几次，经过不断的调整发现我的这种算法的i必须在[104750，104780]范围内。#include#includeusing namespace std;int main(){ int m,n,p[10005]; cin>>m>>n; int i

大概思路：题意是解最少步骤及打印对应步骤。Min求解可以用BFS，但是此题还需记录步骤。    故用BFS+vector解决此题，第一次TLE，加了visited[105][105]判断后就AC了。#include#include#include#includeusing namespace std;struct Pot{ int A,B,steps; ve

Problem DescriptionTechnicians in a pathology lab analyze digitized images of slides. Objects on a slide are selected for analysis by a mouse click on the object. The perimeter of the boundary of

题目链接：http://poj.org/problem?id=1703大概思路：题目要确认的有三种结果，所以不能直接找出a的不同集合再与b合并。由于事先并不知道a、b是哪一派别的，所以开一个2*maxn大的数组，分成两个派别：a属于a，a属于a+maxn。每次执行D操作其实就是合并（a,b+maxn）和（a+maxn,b），最后判断a,b是否同一派别，a,b+maxn是否同一派别即可。TL

注意题目的数据大小，ans要用long long类型。#include #includeusing namespace std;int n,b[200005];void solve();int main(){ cin>>n; for(int i=0;i<n;i++){ cin>>b[i]; } sort(b,b+n); sol

（1）从最左侧s点开始找在s+R范围内的最右侧的e点，即为标记点；（2）再从e点出发找最靠近s+R范围的点，更新s点，回到（1）；#include #includeusing namespace std;int r,n,p[1005];void solve();int main(){ while(cin>>r>>n&&r!=-1||n!=-1){ fo

首尾比较时，主要问题在于它们相等的处理。PE的话请仔细看清输出要求。#include using namespace std;int n;char s[2005],e[2005];void solve();int main(){ cin>>n; for(int i=0;i<n;i++){ cin>>s[i]; } solve(

#include#includeusing namespace std;int main(){ int n; while(cin>>n){ int i,j,t=0,m=0,num=0; for(i=2;i<=n;i++){ bool flag=0; for(j=2;j<=sqrt(i);j++){

#include #includeusing namespace std;int main(){ char str[105]; int num[105],len; while(cin>>str){ int i,j; len=strlen(str); for(i=0;i<len;i++){

题目来源：http://poj.org/problem?id=3259第一个是Bellman-Ford  、第二个是Floyd-Warshall。（一）#include #include#includeusing namespace std;struct edge{ int from,to,cost;}ed[6005];int F,N,M,W;int d[5

题目来源：http://poj.org/problem?id=2139题意：有N只牛，在同一部电影Mi里的Ni只牛它们之间的距离为1，当两只牛不在同一部电影里，有第三只牛分别在那两只牛拍的电影里，则它们的距离为2。要求的是以其中一只牛为起点与其他牛之间的距离之和最短。N#include #include#includeusing namespace std;const int

( 1 一开始开全局数组，MLT；后来用栈pop一次push一次，可循环还得O(n)，TLE；最后的最后，只好找规律。#include #includeusing namespace std;int main(){ int a,b,n,flag[7][7]; while(cin>>a>>b>>n){ if(!a&&!b&&!n) break;

简单的贪心，就是在排序上浪费了点时间，还是对二维数组排序的comp函数没辙，最后用了struct。#include #include#includeusing namespace std;struct Node{ int a,b;}node[1005];bool comp(Node a,Node b);int main(){ int n,m; while

简直就要拜倒在这输出上了~#include #includeusing namespace std;int func(int j);int main(){ int i,j; double e; cout<<"n e\n- -----------\n"; for(i=0;i<10;i++){ e=0; for(j=

题意：求解N^N(1)的最左边的数字。思路：遇到这么大的数，只能求救数学公式了，恰好可以运用取对数的方法实现果的简化。

Digital RootsTime Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 11   Accepted Submission(s) : 4Font: Times New Roman | Verdana | Georgia

大概题意：输入为含有p, q, r, s,  t代表的 五个逻辑变量及K, A, N, C, E代表的五个基本逻辑连接词的字符串，输出即判断是否为永真式。思路：参照http://blog.youkuaiyun.com/jun_sky/article/details/7002084的思路，即五个变量共2^5种情况（00000-11111），再从字符串尾部开始压栈，    分成三种情况考虑1）逻辑变量

题目来源：http://poj.org/problem?id=3009咋一看，求最短路径问题，不假思索地用bfs，但各种MLE、TLE，最后不得不转换思维用dfs+剪枝。#include using namespace std;int n,m;int board[25][25];int sx,sy;int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

DescriptionBessie hears that an extraordinary meteor shower is coming; reports say that these meteors will crash into earth and destroy anything they hit. Anxious for her safety, she vows to find

DescriptionGiven 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 order. The remaining digits can be written do

组成三角形的充要条件：最长边长（1）直接枚举：O(n^3)#include #includeusing namespace std;int main(){ int n,a[105]; while(cin>>n){ for(int i=0;i<n;i++){ cin>>a[i]; } int a

DescriptionA factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5, 6*6. These products are always delivered to customers in the square

DescriptionWe all love recursion! Don't we? Consider a three-parameter recursive function w(a, b, c): if a 1 if a > 20 or b > 20 or c > 20, then w(a, b, c) returns: w(20, 20, 20)

DescriptionIn how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles? Here is a sample tiling of a 2x17 rectangle. InputInput is a sequence of lines, each line co

DescriptionLittle Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes. This is an example of

Problem Description73 88 1 02 7 4 44 5 2 6 5(Figure 1)Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route

Problem DescriptionThe most important part of a GSM network is so called Base Transceiver Station (BTS). These transceivers form the areas called cells (this term gave the name to the cellular phone)

DescriptionA subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = another sequence Z = is a subsequence of X if there exists

DescriptionAn army of ants walk on a horizontal pole of length l cm, each with a constant speed of 1 cm/s. When a walking ant reaches an end of the pole, it immediatelly falls off it. When two ant

题目：每次从n个数里抽取一个数记录后并放回，问是否存在抽取的四个数之和为m的情况。当n下面的算法时间复杂度为O(n^2 log n)。#include #includeusing namespace std;int n,m;int a[1005],b[1000005];bool b_search(int x);void solve();int main(){

DescriptionFJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer

DescriptionThe 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 rectilinear grid of digits paral

DescriptionThere is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent til

题目：给定一个大小为N*M的迷宫。迷宫由通道和墙壁组成，每一步可以向邻接的上下左右四格的通道移动。请求出从起点到终点所需的最小步数。#include #includeusing namespace std;const int INF=100000;int n,m,sx,sy,ex,ey;int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};int step

DescriptionDue to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains

模拟，vector#include #include#includeusing namespace std;int n;vector v[30];void find_block(int a,int& p,int& h);void back_odd(int p,int h);void add_onto(int p1,int h,int p2);int main(){