算法竞赛入门经典第二版第二章习题及思考题

Posted 2021-01-11 163467wyj

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enmmmmmm】大部分好像除了最后一个思考题都很简单

代码如下

#include <iostream>
#include  <cstring>
#include   <cstdio>
#include    <cmath>

int main(){
    /*
    for(int i =100 ; i <= 999 ; i++)
    {
        int a = i/100;int c = i%10;int b = (i-a*100)/10;//水仙花数
        if( i == pow(a,3)+pow(b,3)+pow(c,3)) WA
        //if( i == a*a*a+b*b*b+c*c*c) //AC
            printf("%d=  %d^3 + %d^3 + %d^3
",i,a,b,c);
    }*/
    
    /*int a,b,c,kase=1;//2-2hanxin韩信
    while(scanf("%d%d%d",&a,&b,&c) == 3 )//3 5 7
    {
        int flag =0;
        printf("Case %d: ",kase++);
        a = a*70+b*21+c*15;
        while(a >= 105){
            a -= 105;
        }
        if(a <=10 || a>=100 )printf("No answer
");
        else printf("%d
",a );
        //for(int i = 10;i <= 100 ; i++)
        //    if(i%3==a && i%5==b && i%7==c)
        //        {printf("%d
",i);flag=1;}
        //if(!flag) printf("No answer
" );

    }*/
    /*
    int n;scanf("%d",&n);//2-3triangle倒三角形
    for(int i = n;i >= 1 ;i--)
    {
        for(int j = n-i  ;j > 0;j--)printf(" ");
        for(int j = i*2-1;j > 0;j--)printf("#");
        for(int j = n-i  ;j > 0;j--)printf(" ");
        printf("
");
    }*/
    /*int n,m;while( scanf("%d%d",&n,&m)==2 )
    {//2-4子序列的和 subsequence
        if(n == 0 && n == m)break;
        double ans = 0.0;
        for( int i = n ; i <= m;i++)
        {
            ans += (1.0/i)*(1.0/i); //直接1.0/(i*i)就炸
//sum += 1 / (((double)n + i) * ((double)n + i)); 
//在n较大时，((double)n + i) * ((double)n + i)会溢出，
//所以应该改成 1 / ((double)n + i) / ((double)n + i)
            //printf("i=%d ans=%.5lf
", i,ans);
        }printf("%.5lf
", ans);//printf("%u %u
", -1,-2);
    }*/
    /*

        int a,b,c,n,n1;
        while(scanf("%d%d%d",&a,&b,&c)==3 && (a||b||c)) 
            { 
                for(int count=0; ;count++) 
                    {
                     if(count) 
                         {
                         if(count<=c) 
                            printf("%d",n); 
                        else 
                            { 
                                if(n1=a/b >= 5) n++; 
                                printf("%d",n); 
                                break; 
                            } 
                        n=a/b; a=(a-n*b)*10; 
                        if(count == 1) std::cout<<"."; 
                         }    
                    }
            } 
    int a,b,c,kase=0,i;
    while(scanf("%d%d%d",&a,&b,&c)&&a&&b&&c){
        printf("Case %d: %d.",++kase,a/b);
        a%=b;
        for( i = 1; i < c; i++){
            printf("%d",a*10/b);
            a=a*10%b;
        }
    
        if( a*10%b*10/b >= 5) a*10/b++;
          printf("%d
",a*10/b);
    }*/
    /*
    int a,b,c;//2-5分数化小数 decimal
    while(scanf("%d%d%d",&a,&b,&c)==3 ){
        if( a== b && b==c && a==0)break;
        int aa[505],bb[505];
        aa[0] = a/b;
        aa[1] = a*10/b;

        for(int i=2; i <= c+1;i++)
        {
            bb[i] = a*10%b;
            aa[i] = bb[i]*10/b;
        }//printf("%d.",aa[0]);
        if(aa[c+1] >= 5 )aa[c] += 1;
        for(int i=c; i >= 1 ;i--)
            if(aa[i] == 10) 
                aa[i]-=10,aa[i-1]+1;
        printf("%d.",aa[0] );
        for(int i=1; i < c; i++)
            printf("%d",aa[i] );
        printf("%d
",aa[c]);
    }
    int a,b,c;while(scanf("%d%d%d",&a,&b,&c) == 3 && a){
            //if(a <= 1e+6 && b <= 1e+6 && c <= 200 && a != 0 && b != 0 && c != 0){
            printf("%.*lf
",c,(double)a / (double)b);
        //}    
    }*/
    /*for(int i=123; i <= 333; i++)
    {//排列
        bool aa[10]={false};
        int flag =0 ;
        aa[i/100] = aa[i/10%10] = aa[i%10]=true;
        int j = i*2; int k = i*3;
        aa[j/100] = aa[j/10%10] = aa[j%10]=true;
        aa[k/100] = aa[k/10%10] = aa[k%10]=true;
        for(int d= 1; d <= 9; d++)
        {
            if( aa[d] == false )
                {flag=1;break;}
        }
        if( !flag )
            printf("%d:%d:%d
", i,j,k);
    }
    int n;scanf("%d",&n);
    for(int i = 1; i <= n; i++)//for(int i = 2; i <= n; i+=2)
        printf("%d
", 2*i);*/
    double i = 0;
    for( i = 0; i != 10; i += 0.1)
    
        printf("%f", i);
    return 0;
}