博文

02(2005-08-22 18:33:00)

摘要:#include <iostream.h>
#include <string.h>
int main()
{
    int n,i;
    char s[100];
    while(cin>>n)
    {
        if(n==0) return 0;
        int max=0;
        for(i=0;i<n;i++)
        {
            gets(s);
            if(s[strlen(s)-4]=='u')
            {
                int sum=0;
                int pos=9;
  &nbs......

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jnu(The Circumference of the Circle)(2005-08-22 14:19:00)

摘要:#include <iostream.h>
#define pi 3.141592653589793
#include <math.h>
#include <fstream.h>
ifstream f("circle.in");
#define cin f
#include<iomanip.h>
int main()
{
    //float a,b,c,d,e,f;
    float x1,x2,x3,y1,y2,y3,x,y,r;
    while(cin>>x1>>y1>>x2>>y2>>x3>>y3)
    {
        float k1,k2;
        /*x1=(long)(a*10);
        y1=(long)(b*10);
        x2=(long)(c*10);
        y2=(long)(d*10);
        x3=(long)(e*10);
        y3=(long)(f*10);
        */......

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jnu(lotto)(2005-08-22 13:16:00)

摘要:#include <iostream.h>
#include <fstream.h>
ifstream f("lotto.in");
#define cin f
int main()
{
    int i,j,k,l,m,n;
    int a[12],b,num;
    int flag=0;
    while(cin>>num)
    {
        if(num==0) return 0;
        if (flag==1) cout<<endl;
        for(i=0;i<num;i++)
        cin>>a[i];
        for(i=0;i<=num-6;i++)
        for(j=i+1;j<=num-5;j++)
        for(k=j+1;k<=num-4;k++)
        for(l=k+1;l<=num-3;l++)
      &......

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另类排序(2005-08-22 10:19:00)

摘要:void sort(int a[],int n)
{int i,j,t;
  for(i=0;i<n-1;i++)
    for(j=i+1;j<n;j++)
      if(a[i]<a[j])   {t=a[i];a[i]=a[j];a[j]=t;}
}
main()
  {int aa[10]={1,2,3,4,5,6,7,8,9,10},i;
   sort(&aa[3],5);
   for(i=0;i<10;i++)  printf("%d,",aa[i]);
   printf("\n");
   }
程序运行后的输出结果是
1,2,3,8,7,6,5,4,9,10,

sort(&aa[3],5);
是从aa[3]开始起对数组的5个元素从大到小排序.
这样排序排序我也是第一次碰到. ......

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jnu(Parencodings)(2005-08-21 16:20:00)

摘要:#include <iostream.h>
int main()
{
    int i,t,n,j,k;
    int a[40],b[20];
    int c[40];
    cin>>t;
    for(k=0;k<t;k++)
    {
        int pre;
        int pos=0;
        cin>>n;
        for(j=0;j<n;j++)
        cin>>b[j];
        pre=0;
        for(j=0;j<n;j++)
        {
            for(i=0;i<b[j]-pre;i++)
            {
&nb......

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jnu(Round and Round We Go)2(2005-08-21 15:36:00)

摘要:#include <iostream.h>
#include <string.h>
#include <fstream.h>
ifstream f("c.in");
#define cin f
int main()
{
    char a[61],b[61];
    char c[61][61];
    while(cin>>a)
    {
        int pos;
        int i,j,k;
        int iw;
        int flag=0;
        for(pos=0;pos<strlen(a);pos++)
        {
            for(i=0;i<strlen(a)-pos;i++)
            c[pos][i]=a[i+pos];
            fo......

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jnu(Round and Round We Go) || stu(1018)(2005-08-21 17:37:00)

摘要:#include <iostream.h>
#include <string.h>
#include <fstream.h>
ifstream f("c.in");
#define cin f
int main()
{
    char a[61],b[61];
    char c[61][61];
    while(cin>>a)
    {
        int pos;
        int i,j,k;
        int iw=0;
        int flag=0;
        for(pos=0;pos<strlen(a);pos++)
        {
            for(i=0;i<strlen(a)-pos;i++)
            c[pos][i]=a[i+pos];
            ......

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jnu(Primary Arithmetic)(2005-08-21 15:29:00)

摘要:#include <iostream.h>
#include <string.h>
#include <fstream.h>
ifstream f("h.in");
#define cin f
int main()
{
    char a[12],b[12],c[12];
    while(cin>>a>>b)
    {
        if(strcmp(a,"0")==0 && strcmp(b,"0")==0)
        return 0;
        int flag=0;
        int sum=0;
        int i;
        for(i=0;i<11;i++)
        c[i]='0';
        c[11]='\0';
        for(i=strlen(a)-1;i>=0;i--)
       &n......

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jnu(Easier Done Than Said?)(2005-08-21 15:28:00)

摘要:#include <iostream.h>
#include <string.h>
#include <fstream.h>
ifstream f("e.in");
#define cin f
int fun(char a)
{
    if(a=='a' || a=='e' || a=='i')
    return 1;
    else if(a=='o' || a=='u')
    return 1;
    return 0;
}    
int main()
{
    char a[100];
    int i,j;
    while(cin>>a)
    {
        int flag=0,flag2=0;
        int sum1=0,sum2=0;
        if(strcmp("end",a)==0) return 0;
        if(strlen(a)==1)  goto loop;
        for(i=0;i<strle......

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pku(2262)(2005-08-20 13:24:00)

摘要:#include <iostream.h>
#include <math.h>
int fun(int n)
{
    int i;
    for(i=2;i<=sqrt(n);i++)
    if (n%i==0) return 0;
    return 1;
}    
int main()
{
    int a;
    while(cin>>a)
    {
        if(a==0) return 0;
        int flag=0;
        int b=3;
        while(1)
        {
            if(fun(b) && fun(a-b))
            {
             ......

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