TJU ACMICPC 200931618

TOJ 1833.   The Hardest Problem Ever

简单题，此处gets(s)和cin.getline(s,201)是一样的，puts(s)和cout<<s<<endl是一样的

#include <iostream>
using namespace std;

int main()
{    
    char s[201];
    int i,len;
    gets(s);
    while(strcmp("ENDOFINPUT", s))
    {
        if (strcmp("START", s) && strcmp("END", s))
        {
            len = strlen(s);
            for (i = 0; i < len; i++)
            {
                if (s[i] <= 'Z' && s[i] >= 'A')
                {
                    s[i] -= 5;
                    if (s[i] < 'A')
                        s[i] += 26;
                }
            }
            puts(s);
        }
        gets(s);
    }
    return 0;
}

TOJ 2491.   Hello Kitty

#include <iostream>
using namespace std;
int main()
{
    char s[500];
    int n,m,i,j;
    while(cin>>s)
    {
        if(s[0]=='.')
            break;
        cin>>n;
        m=strlen(s);
        for(i=0;i<m;i++)
        {
            for(j=i;j<m;j++)
                cout<<s[j];
            for(j=1;j<n;j++)
                cout<<s;
            for(j=0;j<i;j++)
                cout<<s[j];
            cout<<endl;
        }
    }

    return 0;
}

TOJ 1920.   Hangover

对于样本比较少的可以用穷举法，简单题

#include <iostream>
using namespace std;
int main()
{    
    float len;    
    double res[278];
    int i;
    res[0]=0;    
    res[1]=0.5;    
    for(i=1;i<278;i++)    
    {        
        res[i+1]=res[i]+(1/(double)(i+2));    
    }
    while(cin>>len,len!=0)
    {
        i=1;
        while(res[i]<=len)
            i++;
        cout<<i<<" card(s)"<<endl;
    }
    return 0;
}

TOJ 1930.   Doubles

简单题

#include <iostream>
using namespace std;
int main()
{        
    int i,j,k,d,t,twice,num[16],end;
    while(cin>>end,end!=-1)
    {
        d=0;
        i=1;num[0]=end;
        while(cin>>t,t!=0)
        {
            num[i++]=t;
        }        
        for(j=0;j<i;j++)
        {
            twice=num[j]*2;
            for(k=0;k<i;k++)
            {
                if(num[k]==twice)
                {d++;break;}
            }
        }
        cout<<d<<endl;
    }
    return 0;
}

TOJ 1701.   Stamps

简单排序题

#include<iostream>
using namespace std;
int main()
{
    int test,all,frd,i,j,stp[1001],temp,sum,k;
    cin>>test;k=1;
    while(test--)
    {
        cin>>all>>frd;
        for(i=0;i<frd;i++)
            cin>>stp[i];
        for(i=0;i<frd-1;i++)
            for(j=i+1;j<frd;j++)
                if(stp[j]>stp[i])
                {temp=stp[j];stp[j]=stp[i];stp[i]=temp;}
        i=0;sum=0;
        while(sum<all)
        {
            if(i==frd)
            {break;}
            sum+=stp[i];
            i++;
        }
        cout<<"Scenario #"<<k++<<":\n";
        if(sum<all) cout<<"impossible"<<endl;
        else        cout<<i<<endl;
        cout<<endl;
    }
    return 0;
}

TOJ 1793.   Assistance Required

数组题

# include <iostream>
using namespace std;
#define  max 50000
int num[100000];
int findnext(int i,int d)
{
    int j,k=0;
    for(j=i+1;j<max;j++)
    {if(num[j]==0)
            k++;
    if(k==d)
        break;}
    return j;
}

int main()
{
    int i,j;
    for(i=2;i<max;i++)
        num[i]=0;
    i=2;
    while(i<max)
    {
        j=i;num[i]=1;
        while(j<max)
        {
            j=findnext(j,i);            
            num[j]=-1;
        }
        i=findnext(i,1);        
    }

    int n;
    while(cin>>n&&n!=0)
    {
        int l=0;
        for(i=2;i<max;i++)
        {
            if(num[i]==1)
                l++;
            if(l==n)
                break;
        }
        cout<<i<<endl;
    }
    return 0;
}

TOJ 1134.   Lotto

方法一：我自己的方法，用六个数组保存，然后从里面选择就行

# include<iostream>
using namespace std;

int main()
{
    int k,i,j,m,n,p,q,a[101],b[101],c[101],d[101],e[101],f[101];
    cin>>k;
    while(1)
    {
        if(k==0)
            break;
        for(i=0;i<k;i++)
        {cin>>a[i];b[i]=a[i];c[i]=a[i];d[i]=a[i];e[i]=a[i];f[i]=a[i];}
        for(i=0;i<k;i++)
            for(j=i+1;j<k;j++)
            {
                if(j!=i)
                {
                    for(m=j+1;m<k;m++)
                    {
                        if(m!=j&&m!=i)
                        {
                            for(n=m+1;n<k;n++)
                            {
                                if(n!=i&&n!=j&&n!=m)
                                {
                                    for(p=n+1;p<k;p++)
                                    {
                                        if(p!=i&&p!=j&&p!=m&&p!=n)
                                        {
                                            for(q=p+1;q<k;q++)
                                            {
                                                if(q!=i&&q!=j&&q!=m&&q!=n&&q!=p)
                                                    cout<<a[i]<<' '<<b[j]<<' '<<c[m]<<' '<<d[n]<<' '<<e[p]<<' '<<f[q]<<endl;
                                            }
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
            cin>>k;
            if(k!=0)
                cout<<endl;
    }
    return 0;
}

方法二：别人的算法，懒得看了，先贴出来

# include<iostream>
# include<algorithm>
using namespace std;
int num[101];
int main()
{
    int n;
    int f=0;
    while(1)
    {
        scanf("%d",&n);
        if(n==0)
            break;
        if(f==0)
            f=1;
        else
            printf("\n");
        int i,j;
        for(i=1;i<=n;i++)
            scanf("%d",&num[i]);
        sort(&num[1],&num[n+1]);
        int t[7];
        for(i=1;i<=6;i++)
            t[i]=num[i];
        while(1)
        {
            for(i=1;i<=6;i++)
            {
                if(i>1)
                    printf(" ");
                printf("%d",t[i]);
            }
            printf("\n");
            if(t[1]==num[n-6+1])
                break;
            i=6;j=n;
            while(1)
            {
                if(t[i]!=num[j])
                    break;
                i--,j--;
            }
            int k;
            for(k=1;k<=n;k++)
                if(num[k]==t[i])
                    break;
            for(j=i;j<=6;j++)
            {
                t[j]=num[k+1];
                k++;
            }
        } 
    }
    return 0;
}