【NOI OpenJudge】【1.3】编程基础之算术表达式与顺序执行

01: A+B问题

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


02: 计算(a+b)*c的值

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


03:计算(a+b)/c的值

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


04: 带余除法

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



05:计算分数的浮点数值

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



06:甲流疫情死亡率

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
//int a, b;
double a, b;
scanf("%lf%lf",&a,&b);
double c = b/a*100;
printf("%.3lf%%",c);
return 0;
}



07: 计算多项式的值

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



08:温度表达转化

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
double x;
scanf("%lf",&x);
double ans = 5*(x-32)/9;
printf("%.5lf",ans);
return 0;
}



09:与圆相关的计算

#include<cstdio>
#include<iostream>
using namespace std;
const double pi=3.14159;
int main(){
double x;
scanf("%lf",&x);
printf("%.4lf %.4lf %.4lf",x*2,2*pi*x,pi*x*x);
return 0;
}



10:计算并联电阻的阻值

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
float a, b;
scanf("%f%f",&a,&b);
float r = 1/(1/a + 1/b);
printf("%.2f",r);
return 0;
}



11:计算浮点数相除的余数

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
double a, b;
scanf("%lf%lf",&a,&b);
int k = 0;
while(a-k*b >= 0){
k++;
}
k--;
printf("%g",a-k*b);
return 0;
}


12: 计算球的体积

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
double x;
double pi = 3.14;
scanf("%lf",&x);
printf("%.2lf",4.0*pi*x*x*x/3);
return 0;
}



13:反向输出一个三位数

#include<cstdio>
#include<iostream>
using namespace std;
int cur;
void go(){
char ch;
if(++cur <= 3){
ch = getchar();
go();
putchar(ch);
}
}
int main(){
go();
return 0;
}



14:大象喝水

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
double h, r;
double pi = 3.14;
scanf("%lf%lf",&h,&r);
r*=0.01, h *= 0.01;
double v = pi*r*r*h;
double ans = 20/(v*1000);
printf("%d",(int)(ans-0.0001)/1+1);
return 0;
}



15:苹果和虫子

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
int n, x, y;
cin>>n>>x>>y;
int b = (int)((y*1.0/x*1.0-0.001)+1);
cout<<n-b<<'\n';
return 0;
}



16:计算线段长度

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
int main(){
double x1, y1, x2, y2;
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
printf("%.3lf\n",sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));
return 0;
}



17:计算三角形面积

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


18:等差数列末项计算

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



19:A*B问题

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



20:计算2的幂

#include<iostream>
using namespace std;
int main(){
int n;  cin>>n;
cout<<(1<<n);
return 0;
}


posted @ 2018-12-09 11:33  gwj1139177410  阅读(...)  评论(...编辑  收藏
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