kongming_acm的专栏

As you known,a positive integer can be written to a form of products of several positive integers(N = x1 * x2 * x3 *.....* xm (xi!=1 (1<=i<=

<br />#include<iostream><br />#include<cstdio><br />#include<algorithm><br />using namespace std;<br />int main()<br />{<br />    int x,y;char ch;<br />    int ci;cin>>ci;<br />    while(ci--)<br />    {<br />        cin>>ch>>x>>y;<br />        int r;<br /

<br />#include<iostream><br />#include<cstdio><br />#include<algorithm><br />using namespace std;<br />int main()<br />{<br />    //freopen("aaa.txt","r",stdin);<br />    //freopen("1.txt","w",stdout);<br />    int x,y;<br />    while(cin>>x>>y)<br />    {

#include#include#includeusing namespace std;bool cmp(string a,string b){    string c=a+b;    string d=b+a;    return c>d;}    int main(){    int n;    string a[60];    while(cin>>n&&n)    {        for(int i=0;i>a[i];        sort(a,a+n,cmp);        for(int

<br />1063: A Funny GameResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K627219Standard<br />1st JOJ Cup Online VContest Warmup Problem<br />Tom and Jack are on the train. The journey is too dull so much so that both of them are eager to find a way

<br />2252: Pick BallsResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K373158StandardSherry is an excellent student of Jilin University. Today her teacher gives her a bag which is filled with M balls. She wants to pick all of them out. To make it i

<br /> 1044: ParencodingsResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K613302Standard<br />Let S = s1 s2 … s2n be a well-formed string of parentheses. S can be encoded in two different ways:By an integer sequence P = p1 p2 … pn where pi is the n

<br />1491: BiorhythmsResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K354147Standard<br />Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical, emotional, and

<br />2236: Balance and Poise<br />ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE5s8192K300133Standard<br />工匠大师Hill打造出一个精致的天平, 对他的徒弟Oren说：“你帮我做一些砝码配套，为了方便携带，砝码数目要尽量少”。 Oren：“好。我将按照2进制的规则打造，也就是1，2，4，8，16等等，这样在称量同等重量下使用的砝码数量最少，只使用7个砝码就可以称量1－127的重量。” <

<br />2428: Super PrimeResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K31388Standard<br />In these days, skywind had thought a series of new problem. But he must keep these problem for the coming important contests such as "JiLin Provincial Colleg

<br />1926: Factovisors<br />ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K659122StandardThe factorial function, n! is defined thus for n a non-negative integer: 0! = 1 n! = n * (n-1)! (n > 0)We say that a divides b if there exists a

<br />#include<stdio.h>   <br />int mod(int c[],int b[],int n)   <br />{   <br />    int all_multy=1,sum=0;   <br />    int i,j,x[5];   <br />    for(i=0;i<n;i++)   <br />        all_multy*=c[i];   <br />    for(i=0;i<n;i++)   <br />        x[i]=all_multy/

<br />1716: DivisorsResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K317113Standard<br />Mathematicians love all sorts of odd properties of numbers. For instance, they consider 945 to be an interesting number, since it is the first odd number for w

<br />#include<stdio.h><br />int main()<br />{<br />    double x1,x2,x3,y1,y2,y3;<br />    while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)==6)<br />    {<br />        double x4=(x1+x2)/2,y4=(y1+y2)/2;<br />        double x5=(x1+x3)/2,y5=(y1+y3)/2

<br />2661: 数字游戏ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE2s32768K35740Standard<br />Raven很喜欢数字游戏，今天他看到两个数，就想能否通过简单的加减，使最终答案等于1。而他又比较厌烦计算，所以他还想知道最少经过多少步加减才能得到1。好吧，我想，是你开动脑筋的时候了。 (注意：至少需要做一步加减法) Input<br />输入包含多组数据。第一行有一个正整数T,表明输入数据的组数。接下来T行，每行有2个非负

<br />#include<iostream><br />#include<cstdio><br />#include<cstring><br />using namespace std;<br />int a[10005],pos[10005];//pos::important 记录位置 <br />int main()<br />{<br />    int  n;char ch;<br />    while(cin>>n&&n)<br />    {<br />        cin>>ch;<b

 1928: Prime DistanceResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K4919StandardThe branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands o

2653: 不同的数ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s65536K38778Standard原来有n对数字

Farmer John is stuck with another problem while feeding his cows. All of his N (1 ≤ N ≤ 100,000) cows (numbered 1..N) are lined up in front

<br />2519: Find the longest sectionResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE1s8192K13329Standard<br />There are n strings with the

给出n,k.求含有n个节点的不同二叉树数目%k InputMULTI TEST CASE!!每行两个整数n,k(1一行二叉树的数目%k Sample Input3 100Sample Output5//师哥模板#include #include #include #inclu

SumTime Limit: 9000 MS Memory Limit: 65536 K SumMr. Jojer is given n numbers and an extra integer x, he wants to know whether the

<br /> 2306: MarriageResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K50373Standard<br />Now, a lot of persons holding their marriages together are in fashion. One day, a lot of people hold their marriages together. They are all happy, so they want

<br /> 2437: The grabberResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K26662Standard<br />When you buy land at the vast grassland, the area depends on your equestrianism. The farmer will give you a horse after received a fixed amount of money. Yo

<br /> 2386: Find the CoinResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K213104Standard<br />One day, Φ has 8 golden coins. He knew one of them is heavier than others. He wants to find out it. Then he wants to borrow X’s scale. But X is so mean.

2701: PartyResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE1s8192K323Standard<br />Go, go, go, it's party time !!! A great many of guests are invited. Guests are numbered from 1 to N. To make the party more joyful, every 3 guests, who are friends to each

<br />2576: The K-th minimum numberResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE5s32768K44138Standard<br />This problem is finding the K-th minimum number of a Matrix. The matrix is defined as below:1. The matrix C has N rows and M columns. 2. C[i][j]

#include#include#includeusing namespace std;const int maxn=101;int mod=1000;class Matrix{    int n,m;//矩阵的row and colum    int num[maxn][maxn];//从1开始    public:        Matrix(){memset(num,0,sizeof(num));}        Matrix(const Matrix &a) {n=a.n;m=a.m; memcpy

<br />2174: AccumulatorResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K324109StandardAccumulator can be used to count the number of the objects. Initially, the value of accumulator is set to zero. After each objected is counted, the value of accum

 2606: 星空ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE1s65536K17918Standardsky是一个喜欢浪漫的人，对浩淼的星空总是有无穷的幻想，每天晚上他都会在独院里听着music，欣赏着银河的璀璨，可是随着时间的的延续，他发现天空中的每颗星的亮度竟都是不同的。而且每一秒都在不停的闪烁，亮度总是不同，真是神奇的自然，他看到天上的n颗星辰，他记得爷爷曾经告诉他星星在第s初时的亮度分别为l[1][0]~l[n][0]，每一秒，在同一时刻的同

<br />1004: Octal FractionsResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K3535790Standard<br />Fractions in octal (base 8) notation can be expressed exactly in decimal notation. For example, 0.75 in octal is 0.963125 (7/8 + 5/64) in decimal. All

<br /> 2475: Matrix TopsResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE5s131072K67978Standard<br />This is an interesting problem about finding the first Z minimum numbers of a Matrix. The matrix is defined as below: <br /> The matrix C has N rows and M

<br />2429: Difference of Square RootResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K1161147Standard<br />At the previous problem , the difference of square is asked. This time, the square root is appeared.<br />Find minimal integer x that satifie

<br /> 2049: Ackermann's FunctionResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K928273Standard<br />In this easy problem, you are to write a program to calculate the Ackermann's Function A:<br />A(0, n) = n+1<br />A(m, 0) = A(m-1, 1)<br />A(m, n)

<br />#include<iostream><br />using namespace std;<br />int main()<br />{<br />    int n;<br />    while(scanf("%d",&n)==1&&n)<br />    {<br />        int sum2=0,sum3=0;<br />        for(int i=2;i<=n;i++)<br />        {<br />            int t=i;<br />     

<br />#include<stdio.h><br />const int inf=999999;<br />int gcd (int x,int y)<br />{<br />    if(y==0) return x;<br />    return gcd(y,x%y);<br />}<br />int main()<br />{<br />    int n,i,j;<br />    //freopen("out.txt","w",stdout);<br />    for(n=1;n<inf;

<br />#include<iostream><br />#include<algorithm><br />using namespace std;<br />int main()<br />{<br />    int n,m,s;<br />    while(scanf("%d%d",&n,&m),n||m)<br />    {<br />  if(n==1||m==1) {printf("0/n");continue;}<br />        if(n%2==1&&m%2==1)<br />

<br />#include<stdio.h><br />int main()<br />{<br /> int n,m;<br /> while(scanf("%d%d",&n,&m)==2&&n)<br /> {<br />  int i,count=0;<br />  for(i=n-m+1;i<=n;i++)<br />  {<br />   int b=i;<br />   while(b%2==0) <br />   {<br />    b/=2;<br />    count++;<br /

<br /> 2544: Length of RopeResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE1s32768K17458Standard<br />There is a convex polygon whose vertices are all nails. The nails are of radius R. We need a rope to round these nails tightly. You can see an example h

<br />2285: DecomposeResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K267149StandardGive you an positive integer N(1<=N<=30), you can decompose n to several positive integers: A1, A2, ... Ak. (1<=k<=N) And A1 + A2 + ... + Ak = N. Now i want to know