//输入2个整数num1和num2.计算并输出它们的和，差，积，商，余数。//

#include<stdio.h>
int main(void)
{int num1,num2;        int x,y,z;           //x为和，y为差，z为积，d为商，e为余数//
    double d,e;printf("输入num1，num2:");scanf("%d",&num1);scanf("%d",&num2);x=num1+num2;y=num1-num2;z=num1*num2;d=num1/num2;e=num1%num2;printf("x=num1+num2=%d\n",x);        //输出结果//
    printf("y=num1-num2=%d\n",y);printf("z=num1*num2=%d\n",z);printf("d=num1/num2=%.2f\n",d);printf("e=num1%num2=%.2f\n",e);return 0;
}

//输入实数x和正整数n，计算x的n次幂的值，不允许调用pow函数求幂//

#include<stdio.h>
int main(void)
{int i,n;double x,y;y=1;                      //赋初值y=1//
    printf("Enter x:");scanf("%lf",&x);printf("Enter n:");scanf("%d",&n);for(i=1;i<=n;i++){         //使用循环语句计算x的n次幂的值//
        y=y*x;}printf("y=%.2f\n",y);return 0;
}

//输入2个正整数m和n，求i*i+1/i 从i等于m开始//

#include<stdio.h>
int main(void)
{int m,n;double sum,i;sum=0;                       //赋初值为0//
    printf("enter m,n:");scanf("%d%d",&m,&n);for(i=m;i<=n;i++){            //当i小于等于n时，持续累加//
        sum=sum+i*i+1/i;}printf("sum=%.2f\n",sum);return 0;
}

//输入一个正整数n，计算1-2/3+3/5-4/7+5/9-6/11+...的前n项之和//

#include<stdio.h>
int main(void)
{double n,i,x,y,w;          //x为第i项的符号,y为第i项的分母//
    double sum,z;          //sum为累加的和,z为第i项的值//
    printf("Enter n:");scanf("%lf",&n);x=1;               //x为第i项的符号,初始为正//
    y=1;                 //y为第i项的分母，初值为1//   w=1;                 //w为第i项的分子，初值为1//
    sum=0;              //置累加和sum的初值为0//
for(i=1;i<=n;i++){z=x*(w/y);            //计算第i项的值//
        sum=sum+z;              //累加和//
        x=-x;               //改变符号，为下一次循环做准备//
        y=y+2;               //分母递增2，为下一次循环做准备//
        w=w+1;               //分子递增1，为下一次循环做准备//
        }printf("sum=%.2f\n",sum);return 0;
}

//输入一个正整数n，求2^i的和，可以调用pow函数求幂//

#include<stdio.h>
#include<math.h>           //调用pow函数//
int main(void)
{int n,i;               double x,sum;          //x为第i的值，sum为和//
    printf("Enter n:");scanf("%d",&n);sum=0;                     //赋初值//
for(i=1;i<=n;i++){x=pow(2,i);           //计算第i项的值//
        sum=sum+x;            //累加和//
    }printf("sum=%.2f\n",sum);return 0;
}