代码改变世界，keep coding。

点击打开链接Second-price AuctionCrawling in process...Crawling failedTime Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%lld & %lluSubmitStatusDescriptionDo

点击打开链接Max SumTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 98077    Accepted Submission(s): 22598Problem Descrip

点击打开链接 Mad Scientist   Time Limit: 1000MS    Memory limit: 65536K 题目描述  A mad scientist performed a series of experiments, each having n phases. During each phase, a measurement was take

点击打开链接  Multitasking Calendars control our daily lives. For people like me, who are bad at multitasking, it is important to have at most one task planned for any minu

点击打开链接  Problem F: Magic MultipleTime Limit: 1 Sec  Memory Limit: 128 MBSubmit: 36  Solved: 23[Submit][Status][Web Board]DescriptionThe Elvish races of Middle Earth believed that

不解释。#includeint main(){ int m; double a,b,c; scanf("%d",&m); while(m--) { scanf("%lf%lf%lf",&a,&b,&c); if(a+b>c&&a+c>b&&b+c>a) printf("YES\n"); else

不解释。#include#includechar z[107];int main(){ while(gets(z)) { int max=z[0]; for(int a=0;a<strlen(z);a++) if(z[a]>max){max=z[a];} for(int b=0;b<strlen(z);b++) if(z[b]

就因为少了个{}，找了半个小时错，以后要仔细。#include#include#includeusing namespace std;char z[57];int main(){ int m; while(scanf("%d",&m)!=EOF) { getchar(); while(m--) {

不解释。#includeint p[107];int zhang[1007];int main(){ int m; while(scanf("%d",&m)!=EOF&&(m!=0)) { for(int a=1; a<=m; a++) { scanf("%d",&p[a]); in

不解释.#include#includeint z[1007];int main(){ int m,flag; while(scanf("%d",&m)&&m!=0) { for(int a=0;a<m;a++) scanf("%d",&z[a]); for(int b=0;b<m-1;b++) f

不解释。#includeint t[100001]={0,2};int main(){ for(int b=2;b<10001;b++) t[b]=t[b-1]+6*(b-1); int n,m; scanf("%d",&n); for(int a=0;a<n;a++) {scanf("%d",&m); printf("%d\n",t[

不解释。#include __int64 a[100005];int main(){ for(int i=1; i<=100000; i++) a[i]=i*i%10000; int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n);

不解释。#include#includechar z[107];int main(){ int n; scanf("%d",&n); while(n--) { int t; scanf("%d",&t); int m=3*t; while(m--) { for(int b=0;b<t;b++)

不解释。#include#includechar z[107];int main(){ while(gets(z)) { z[0]=z[0]-32;printf("%c",z[0]); for(int a=1; a<strlen(z); a++) { if(z[a-1]==' '&&z[a]!='

不解释。#include#includechar z[10007];int main(){ int n; while (scanf("%d",&n)!=EOF) { int count=0; for (int a=0;a<n;a++) { scanf("%s",&z);

不解释。#include int z[107];int main(){ int temp,n; while (scanf("%d",&n),n) { int min; for (int a=0;a<n;a++) { scanf("%d",&z[a]); } min=

不解释。#include double z[107];int main(){ int x; while (scanf("%d",&x)!=EOF) { int y; for (int a=0;a<x;a++) { double ans=0,m; scanf("%d",&y)

不解释。#include int z[107];int main(){ int x; while (scanf("%d",&x)!=EOF) { int min=100,max=0; double count=0; for (int a=0;a<x;a++) { scanf("%d

不解释。#include #include int main(){ double n,m; while(scanf("%lf%lf",&n,&m)!=EOF) { double x=0,y=n; for(int a=1;a<m;a++) { x=sqrt(n); y+=x; n=x; } printf("%.2f\n",y); } return

#include #define pi 3.1415927int main(){ double r; while (scanf("%lf",&r)!=EOF) { double s; s=4*pi*r*r*r/3; printf("%.3f\n",s); } return 0;}

不解释。#include int main(){ int m; while (scanf("%d",&m)!=EOF&&m!=0) { double n; int fu=0,zheng=0,zero=0; for(int a=0;a<m;a++) { scanf("%lf",&n); i

只因为一个if条件不小心加{}里面，把我弄郁闷了好几个小时，真是编程的孩子伤不起啊。#includeint z[50][5];int main(){ int m,n; while(scanf("%d%d",&n,&m)!=EOF) { int flag=0,k=0,count=0; double fen=0,ren=0,grad

不解释。#include#includechar z[107];int main(){ int n; scanf("%d",&n); getchar(); for(int c=0; c<n-1; c++) { gets(z); int y1=0,y2=0,y3=0,y4=0,y5=0; for(i

不解释。#include #include int A(int n){ return n*n+n+41; }int main(){ int x,y,flag=0; while(scanf("%d%d",&x,&y)!=EOF&&(x!=0||y!=0)) { int z,count=0; for(int n=x;n<=y;n++) {

不解释。#include int y[1007];int main(){ int m,n; while (scanf("%d%d",&m,&n)!=EOF) { int bai,shi,ge,z,x=0,count=0; for (int a=m;a<=n;a++) { bai=a/100;

不解释。#include int main(){ int m,n; while (scanf("%d%d",&m,&n)!=EOF) { int temp,j=0,o=0; if (m>n) { temp=m; m=n; n=temp;

不解释。#include int z[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};int main(){ int n,y,r; while(scanf("%d/%d/%d",&n,&y,&r)!=EOF) { int t=0; if((n%4==0&&n%100!=0)||(n%400==0)) z[2]=29;

不解释。#include int main(){ int n,x,y,temp,z,count=0; while(scanf("%d",&n)!=EOF) { while(n--) { scanf("%d%d",&x,&y); if(x>y) { temp=x; x=y; y=temp; } for(int a=x;a<=

点击打开链接男生女生配Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^题目描述 学校有A个同学参加数学竞赛，如果将参赛人员分成B组，则必有一组的女生的女生的人数大于C，又知参赛者中任意D人中至少有男生E人，则参赛的男生最多有Y人，最少有X人。如果结果逻辑错误，输出-1.

点击打开链接画8Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3300    Accepted Submission(s): 1461Problem Description谁画8画的好,画的快,今后

点击打开链接Quoit DesignTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24618    Accepted Submission(s): 6542Problem Descriptio

点击打开链接Problem 2122 又见LKityAccept: 76    Submit: 266Time Limit: 1000 mSec    Memory Limit : 32768 KB Problem Description嗨！大家好，在TempleRun中大家都认识我了吧。我是又笨又穷的猫猫LKity。很高兴这次又与各位FZU的A

点击打开链接Problem 2027 单词问题Accept: 335    Submit: 1230Time Limit: 1000 mSec    Memory Limit : 32768 KB Problem Description给出一个完整的句子，这个句子中不包含不可见字符或者空格，于是在这个句子中有许多不同的单词。一个单词是指一串连续的最长

点击打开链接Simplest Task in WindowsTime Limit: 2 Seconds      Memory Limit: 65536 KBA typical windows platform may have several windows on the desktop. A user's operation may be as simple a

点击打开链接 Problem 1055 赋值问题Accept: 1386    Submit: 7564Time Limit: 1000 mSec    Memory Limit : 32768 KB Problem Description在很多程序设计语言中，忘记给变量赋初值的错误常令人头疼。请编程求出含N(0≤N≤100)行的程序段运行以后有

点击打开链接A Stack or A Queue?Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%lld & %lluSubmitStatusDescriptionDo you

这是一道地推的题目，算几个数你就会发现这是一个斐波那契，然后就好做了。#includeint z[50]={0,1,1};int main(){ for(int a=3;a<50;a++) z[a]=z[a-1]+z[a-2]; int m; scanf("%d",&m); while(m--) { int n;

一道比较简单的递推问题。#includeint main(){ int z[31]= {0,4}; for(int a=2; a<=31; a++) z[a]=(z[a-1]-1)*2; int n; scanf("%d",&n); while(n--) { int m; scanf("%d",&

点击打开链接  选夫婿1 Time Limit: 1000MS    Memory limit: 32768K 题目描述    倾国倾城的大家闺秀潘小姐要选夫婿啦！武林中各门各派，武林外各大户人家，闻讯纷纷前来，强势围观。前来参与竞选的男生藏龙卧虎，高手云集，才子遍布，帅哥纷纭，更不乏富二代，官二代，可谓声势空前。    每个人参与竞选的帅哥除了进行一段

点击打开链接   排序问题 Time Limit: 1000MS    Memory limit: 65536K 题目描述输入10个整数，将它们从小到大排序后输出，并给出现在每个元素在原来序列中的位置。输入输入数据有一行，包含10个整数，用空格分开。输出输出数据有两行，第一行为排序后的序列，第二行为排序后各个元素在原来序列中的位置。示例输入