紫书第二章习题

2-1

#include <iostream>

2-2

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#define INF 1000000000
using namespace std;

int main()
{
   int a,b,c;
   int kase = 1;
   while(cin>>a>>b>>c){
    int ans = 0;
    ans = a*70 + b*21 + c*15;
    while(ans > 105)
        ans -= 105;
    if(ans > 100 || ans < 0 )
        cout<<"Case "<<kase<<": "<<"No answer"<<endl;
    else
    cout<<"Case "<<kase<<": "<<ans<<endl;
    kase++;
   }
    return 0;
}

2-3

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#define INF 1000000000
using namespace std;

int main()
{
    int n;
    while(cin>>n){
        for(int i = n;i > 0;i--){
            for(int j = 0;j < n-i;j++)
                cout<<" ";
                for(int k = 0;k < 2*i-1;k++){
                    cout<<"#";
                }
            cout<<endl;
            }

    }
    return 0;
}

2-4

//本题陷阱在于n比较大时，n*n会溢出，所以 1/n^2 应该用 1/n/n 而不是 1/(n*n)。
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#define INF 1000000000
using namespace std;

int main()
{
    int n,m;
    int kase = 1;
    while(cin>>n>>m && (n||m)){
        double ans = 0;
        for(int i = n;i <= m;i++){
            ans += (double)1/i/i;
        }
        printf("Case %d: %.5f\n",kase,ans);
        kase++;
    }
    return 0;
}

2-5

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <iomanip>
#define INF 1000000000
using namespace std;

int main()
{
   int a,b,c;
   int kase = 0;
   while(cin>>a>>b>>c &&(a||b||c)){
    //cout << fixed << setprecision(c)<<(double)a/b<<endl;//不用那么麻烦的做法
    kase++;
        printf("Case %d: %d.",kase,a/b);//打印整数部分及小数点
        a%=b;
        for(int i=1;i<c;i++){
            printf("%d",a*10/b);
            a=a*10%b;
        }
        //最后一位，四舍五入处理
        if(a*10%b*10/b>=5)//c+1位数据判断，若大于等于5
            printf("%d\n",a*10/b+1);//c位需 五入
        else
            printf("%d\n",a*10/b);
   }
    return 0;
}

2-6

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#define INF 1000000000
using namespace std;
int main()
{

    int def,ghi;
    int a,b,c,d,e,f,g,h,i;
    for(int abc = 123;abc < 333;abc++){
        bool arr[10] = {false};
        a = abc/100;
        b = abc/10%10;
        c = abc%10;
        if(a == 0 || b == 0 || c == 0 || a == b || b == c || a == c) continue;
        arr[a] = arr[b] = arr[c] = true;
        def = 2*abc;
        d = def/100;
        e = def/10%10;
        f = def%10;
        if(d == e || e == f || d == f) continue;
        if(arr[d] == false && arr[e] == false && arr[f] == false)
            arr[d] = arr[e] = arr[f] = true;
        else continue;
        ghi = 3*abc;
        g = ghi/100;
        h = ghi/10%10;
        i = ghi%10;
        if(arr[g] == false && arr[h] == false && arr[i] == false)
            printf("%d %d %d\n",abc,def,ghi);
        else continue;
    }
    return 0;
}
思考题2

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#define INF 1000000000
using namespace std;
int main()
{

    double i;
    for(i = 0;i != 10;i += 0.1)
        printf("%.1f\n",i);
    return 0;
}
//无限循环