1.3编程基础之算术表达式与顺序执行

刷水题，刷水题

Portal：http://noi.openjudge.cn/ch0103/

01 a+b

02 (a+b)*c

03 (a+b)/c

04 带余除法

05 计算分数的浮点数值

都是看题意能看懂的题

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    cout<<a+b;
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
int a,b,c;
cin>>a>>b>>c;
cout<<(a+b)*c;
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
int a,b,c;
cin>>a>>b>>c;
cout<<(a+b)/c;
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
int a,b;
cin>>a>>b;
cout<<a/b<<‘ ‘<<a%b;
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
double a,b;
cin>>a>>b;
printf("%.9lf",a/b);
    return 0;
}

06 http://noi.openjudge.cn/ch0103/06/

意外的收获：输出%可以使用printf(“%%”); 233 那么 输出”可以使用printf(“””); 233

07 计算多项式的值

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
double a,b,c,d,x;
cin>>x>>a>>b>>c>>d;
printf("%.7lf",a*x*x*x+b*x*x+c*x+d);
    return 0;
}

其实直接用pow好一点？

过几天搞搞快速多项式除法试试?

08-20

水题，少部分有坑点

直接贴代码

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
double f;
cin>>f;
printf("%.5lf",5*(f-32)/9);
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
double r;
cin>>r;
printf("%.4lf %.4lf %.4lf",r*2,r*2*3.14159,r*r*3.14159);
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
float r1,r2;
cin>>r1>>r2;
printf("%.2lf",1/(1/r1+1/r2));
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
double a,b;
cin>>a>>b;
int c;
c=a/b;
printf("%g",a-b*c);
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
double r;
cin>>r;
printf("%.2lf",3.14*r*r*r*4/3);
    return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
int n;
cin>>n;
cout<<n%10<<(n/10)%10<<n/100;
return 0;
}

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int v,h,r;
    cin>>h>>r;
    v=h*r*r*3.14;
    cout<<20000/v+1;
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    double xa,ya,xb,yb;
    cin>>xa>>xb>>ya>>yb;
    printf("%.3lf",sqrt((xa-ya)*(xa-ya)+(xb-yb)*(xb-yb)));
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    double a,b,c,r;
    cin>>a>>b>>c;
    if ((a+b>c)&&(c+b>a)&&(a+c>b))
    {
        r=(a+b+c)/2;
        printf("%.4lf",sqrt(r*(r-a)*(r-b)*(r-c)));
    }
    else cout<<"Data Error";
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    double x1,x2,x3,y1,y2,y3,a,b,c,r;
    cin>>x1>>y1>>x2>>y2>>x3>>y3;
    a=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
    b=sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));
    c=sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));
        r=(a+b+c)/2;
        printf("%.2lf",sqrt(r*(r-a)*(r-b)*(r-c)));
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int a1,a2,n;
    cin>>a1>>a2>>n;
    cout<<a1+(a2-a1)*(n-1);
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
long long a,b;
cin>>a>>b;
cout<<a*b;
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
#define FOR(i,j,k) for(int i=j;i<=k;i++)
int main()
{
long long a,b;
cin>>a;
b=1;
FOR(i,1,a)
b=b*2;
cout<<b;
    return 0;
}

意外的收获：海伦公式