‘壹’ 用c语言编程，求一个公式的和，下面有我的代码以及问题，求大神指教
你的代码问题出在求组合上，把对a、b的求解部分代码：
h=(i>n+1-i)?i:n+1-i;
......
s=s+a/b*t;
其中h、k的赋值显得太麻烦。换成如下代码：
a=b=1;
for(j=i,h=n+1-i,k=i;j>0;j--,h--,k--){//执行i次连乘
a=a*h;
b=b*j;
}
s=s+a/b*t;
通过g++编译运行了，结果为，
n=1 输出3
n=2输出8
n=3输出22
......
n=40输出309883934660886528
‘贰’ c语言 编程求以下公式的值 s=1+1/(1*2)+1(2*3)+...1/(n*(n+1))
voidmain()
{
floatsum;
intn;
again: printf("请输入n的值 ");
scanf("%d",&n);
sum=1.0;
for(inti=1;i<=n;i++)
sum+=1.0/(i*(i+1));
printf("sum=%f ",sum);
gotoagain;
}
测试无误，绝对精确。
‘叁’ C语言编一个实现简单的算术表达式求值的代码。
这个是个简单的学生管理系统,我们学校的程序大作业就是做这个...呵呵,希望能帮到你
#include<stdio.h> /*引用库函数*/
#include<stdlib.h>
#include<ctype.h>
#include<string.h>
typedef struct /*定义结构体数组*/
{
char num[10]; /*学号*/
char name[20]; /*姓名*/
int score; /*成绩*/
}Student;
Student stu[80]; /*结构体数组变量*/
int menu_select() /*菜单函数*/
{
char c;
do{
system("cls"); /*运行前清屏*/
printf("\t\t****Students' Grade Management System****\n"); /*菜单选择*/
printf("\t\t | 1. Input Records |\n");
printf("\t\t | 2. Display All Records |\n");
printf("\t\t | 3. Sort |\n");
printf("\t\t | 4. Insert a Record |\n");
printf("\t\t | 5. Delete a Record |\n");
printf("\t\t | 6. Query |\n");
printf("\t\t | 7. Statistic |\n");
printf("\t\t | 8. Add Records from a Text File|\n");
printf("\t\t | 9. Write to a Text file |\n");
printf("\t\t | 0. Quit |\n");
printf("\t\t*****************************************\n");
printf("\t\t\tGive your Choice(0-9):");
c=getchar(); /*读入选择*/
}while(c<'0'||c>'9');
return(c-'0'); /*返回选择*/
}
int Input(Student stud[],int n) /*输入若干条记录*/
{int i=0;
char sign,x[10]; /*x[10]为清除多余的数据所用*/
while(sign!='n'&&sign!='N') /*判断*/
{ printf("\t\t\tstudent's num:"); /*交互输入*/
scanf("\t\t\t%s",stud[n+i].num);
printf("\t\t\tstudent's name:");
scanf("\t\t\t%s",stud[n+i].name);
printf("\t\t\tstudent's score:");
scanf("\t\t\t%d",&stud[n+i].score);
gets(x); /*清除多余的输入*/
printf("\t\t\tany more records?(Y/N)");
scanf("\t\t\t%c",&sign); /*输入判断*/
i++;
}
return(n+i);
}
void Display(Student stud[],int n) /*显示所有记录*/
{
int i;
printf("\t\t\t-----------------------------------\n"); /*格式头*/
printf("\t\t\tnumber name score\n");
printf("\t\t\t-----------------------------------\n");
for(i=1;i<n+1;i++) /*循环输入*/
{
printf("\t\t\t%-16s%-15s%d\n",stud[i-1].num,stud[i-1].name,stud[i-1].score);
if(i>1&&i%10==0) /*每十个暂停*/
{printf("\t\t\t-----------------------------------\n"); /*格式*/
printf("\t\t\t");
system("pause");
printf("\t\t\t-----------------------------------\n");
}
}
printf("\t\t\t");
system("pause");
}
void Sort_by_num(Student stud[],int n) /*按学号排序*/
{ int i,j,*p,*q,s;
char t[10];
for(i=0;i<n-1;i++) /*冒泡法排序*/
for(j=0;j<n-1-i;j++)
if(strcmp(stud[j].num,stud[j+1].num)>0)
{strcpy(t,stud[j+1].num);
strcpy(stud[j+1].num,stud[j].num);
strcpy(stud[j].num,t);
strcpy(t,stud[j+1].name);
strcpy(stud[j+1].name,stud[j].name);
strcpy(stud[j].name,t);
p=&stud[j+1].score;
q=&stud[j].score;
s=*p;
*p=*q;
*q=s;
}
}
int Insert_a_record(Student stud[],int n) /*插入一条记录*/
{char x[10]; /*清除多余输入所用*/
printf("\t\t\tstudent's num:"); /*交互式输入*/
scanf("\t\t\t%s",stud[n].num);
printf("\t\t\tstudent's name:");
scanf("\t\t\t%s",stud[n].name);
printf("\t\t\tstudent's score:");
scanf("\t\t\t%d",&stud[n].score);
gets(x);
n++;
Sort_by_num(stud,n); /*调用排序函数*/
printf("\t\t\tInsert Successed!\n"); /*返回成功信息*/
return(n);
}
int Delete_a_record(Student stud[],int n) /*按姓名查找，删除一条记录*/
{ char s[20];
int i=0,j;
printf("\t\t\ttell me his(her) name:"); /*交互式问寻*/
scanf("%s",s);
while(strcmp(stud[i].name,s)!=0&&i<n) i++; /*查找判断*/
if(i==n)
{ printf("\t\t\tnot find!\n"); /*返回失败信息*/
return(n);
}
for(j=i;j<n-1;j++) /*删除操作*/
{
strcpy(stud[j].num,stud[j+1].num);
strcpy(stud[j].name,stud[j+1].name);
stud[j].score=stud[j+1].score;
}
printf("\t\t\tDelete Successed!\n"); /*返回成功信息*/
return(n-1);
}
void Query_a_record(Student stud[],int n) /*查找并显示一个记录*/
{ char s[20];
int i=0;
printf("\t\t\tinput his(her) name:"); /*交互式输入*/
scanf("\t\t\t%s",s);
while(strcmp(stud[i].name,s)!=0&&i<n) i++; /*查找判断*/
if(i==n)
{ printf("\t\t\tnot find!\n"); /*输入失败信息*/
return;
}
printf("\t\t\this(her) number:%s\n",stud[i].num); /*输出该学生信息*/
printf("\t\t\this(her) score:%d\n",stud[i].score);
}
void Statistic(Student stud[],int n) /*新增功能，输出统计信息*/
{ int i,j=0,k=0,sum=0;
float aver; /*成绩平均值*/
for(i=0;i<n;i++) /*循环输入判断*/
{
sum+=stud[i].score;
if(stud[j].score>stud[i].score) j=i;
if(stud[k].score<stud[i].score) k=i;
}
aver=1.0*sum/n;
printf("\t\t\tthere are %d records.\n",n); /*总共记录数*/
printf("\t\t\tthe hignest score:\n"); /*最高分*/
printf("\t\t\tnumber:%s name:%s score:%d\n",stud[j].num,stud[j].name,stud[j].score);
printf("\t\t\tthe lowest score:\n"); /*最低分*/
printf("\t\t\tnumber:%s name:%s score:%d\n",stud[k].num,stud[k].name,stud[k].score);
printf("\t\t\tthe average score is %5.2f\n",aver); /*平均分*/
}
int AddfromText(Student stud[],int n) /*从文件中读入数据*/
{ int i=0,num;
FILE *fp; /*定义文件指针*/
char filename[20]; /*定义文件名*/
printf("\t\t\tInput the filename:");
scanf("\t\t\t%s",filename); /*输入文件名*/
if((fp=fopen(filename,"rb"))==NULL) /*打开文件*/
{ printf("\t\t\tcann't open the file\n"); /*打开失败信息*/
printf("\t\t\t");
system("pause");
return(n);
}
fscanf(fp,"%d",&num); /*读入总记录量*/
while(i<num) /*循环读入数据*/
{
fscanf(fp,"%s%s%d",stud[n+i].num,stud[n+i].name,&stud[n+i].score);
i++;
}
n+=num;
fclose(fp); /*关闭文件*/
printf("\t\t\tSuccessed!\n");
printf("\t\t\t");
system("pause");
return(n);
}
void WritetoText(Student stud[],int n) /*将所有记录写入文件*/
{
int i=0;
FILE *fp; /*定义文件指针*/
char filename[20]; /*定义文件名*/
printf("\t\t\tWrite Records to a Text File\n"); /*输入文件名*/
printf("\t\t\tInput the filename:");
scanf("\t\t\t%s",filename);
if((fp=fopen(filename,"w"))==NULL) /*打开文件*/
{
printf("\t\t\tcann't open the file\n");
system("pause");
return;
}
fprintf(fp,"%d\n",n); /*循环写入数据*/
while(i<n)
{
fprintf(fp,"%-16s%-15s%d\n",stud[i].num,stud[i].name,stud[i].score);
i++;
}
fclose(fp); /*关闭文件*/
printf("Successed!\n"); /*返回成功信息*/
}
void main() /*主函数*/
{
int n=0;
for(;;)
{
switch(menu_select()) /*选择判断*/
{
case 1:
printf("\t\t\tInput Records\n"); /*输入若干条记录*/
n=Input(stu,n);
break;
case 2:
printf("\t\t\tDisplay All Records\n"); /*显示所有记录*/
Display(stu,n);
break;
case 3:
printf("\t\t\tSort\n");
Sort_by_num(stu,n); /*按学号排序*/
printf("\t\t\tSort Suceessed!\n");
printf("\t\t\t");
system("pause");
break;
case 4:
printf("\t\t\tInsert a Record\n");
n=Insert_a_record(stu,n); /*插入一条记录*/
printf("\t\t\t");
system("pause");
break;
case 5:
printf("\t\t\tDelete a Record\n");
n=Delete_a_record(stu,n); /*按姓名查找，删除一条记录*/
printf("\t\t\t");
system("pause");
break;
case 6:
printf("\t\t\tQuery\n");
Query_a_record(stu,n); /*查找并显示一个记录*/
printf("\t\t\t");
system("pause");
break;
case 7:
printf("\t\t\tStatistic\n");
Statistic(stu,n); /*新增功能，输出统计信息*/
printf("\t\t\t");
system("pause");
break;
case 8:
printf("\t\t\tAdd Records from a Text File\n");
n=AddfromText(stu,n); /*新增功能，输出统计信息*/
break;
case 9:
printf("\t\t\tWrite to a Text file\n");
WritetoText(stu,n); /*循环写入数据*/
printf("\t\t\t");
system("pause");
break;
case 0:
printf("\t\t\tHave a Good Luck,Bye-bye!\n"); /*结束程序*/
printf("\t\t\t");
system("pause");
exit(0);
}
}
}
‘肆’ 大佬求解c语言编程 求以下公式的值 s=1+1/（1*2）+1（2*3）+…1/（n*（n+1）
#include<stdio.h>
voidmain()
{
inti,n;
doubles=1;
scanf("%d",&n);
for(i=1;i<=n;i++)
s+=1.0/(i*(i+1));
printf("s=%f",s);
}
‘伍’ 怎样用C语言编写数学公式
1、C语言有现场的常用数学函数，所在函数库为math.h、stdlib.h。
函数名及解释：
int abs(int i) 返回整型参数i的绝对值
double cabs(struct complex znum) 返回复数znum的绝对值
double fabs(double x) 返回双精度参数x的绝对值
long labs(long n) 返回长整型参数n的绝对值
double exp(double x) 返回指数函数ex的值
double frexp(double value,int *eptr) 返回value=x*2n中x的值,n存贮在eptr中
double ldexp(double value,int exp); 返回value*2exp的值
double log(double x) 返回logex的值
double log10(double x) 返回log10x的值
double pow(double x,double y) 返回xy的值
double pow10(int p) 返回10p的值
double sqrt(double x) 返回+√x的值
double acos(double x) 返回x的反余弦cos-1(x)值,x为弧度
double asin(double x) 返回x的反正弦sin-1(x)值,x为弧度
double atan(double x) 返回x的反正切tan-1(x)值,x为弧度
double atan2(double y,double x) 返回y/x的反正切tan-1(x)值,y的x为弧度
double cos(double x) 返回x的余弦cos(x)值,x为弧度
double sin(double x) 返回x的正弦sin(x)值,x为弧度
double tan(double x) 返回x的正切tan(x)值,x为弧度
double cosh(double x) 返回x的双曲余弦cosh(x)值,x为弧度
double sinh(double x) 返回x的双曲正弦sinh(x)值,x为弧度
double tanh(double x) 返回x的双曲正切tanh(x)值,x为弧度
double hypot(double x,double y) 返回直角三角形斜边的长度(z),x和y为直角边的长度,z2=x2+y2
2、复杂的数学函数可以用自定义函数的形式实现。
例如：
doubleConerVelocity(intA,doublex1,doubley1,doublex2,doubley2,doublet1,doublet2)
{
doubleT,V;
T=acos(abs(x1*x2+y1*y2)/sqrt(x1*x1+y1*y1)/sqrt(x2*x2+y2*y2));
V=(A*(t2-t1))/(2*sin(T/2));
returnV;
}
‘陆’ c语言表达式求值代码
#include<stdio.h>
#include<malloc.h>
#include<string.h>
#include<stdlib.h>
static int gv_a;
typedef struct list //创建队列元素型
{
char str[10];
struct list*pNext;
}LIST;
typedef struct queue //创建队列表型
{
LIST *front;
LIST *rear;
}QUEUE;
typedef struct oprstack //创建运算符栈型
{
char opr[100];
int a;
}OPRSTACK;
typedef struct valstack //创建运算栈型
{
float val[30];
int a;
}VALSTACK;
typedef struct element //创建表达式栈元素型
{
char c;
struct element *pNext;
}ENT;
typedef struct stack //创建表达式栈 型
{
ENT *pTop;
ENT *bottom;
}STACK;
bool calc(char *str,float *result);
float evaluationofexpression(char *str);
int main(void)
{
char str[50];
float result;
printf("请输入要计算的表达式：");
scanf("%s",str);
result = evaluationofexpression(str);
if (gv_a)
return 0;
printf("结果：%f",result);
printf("\n");
return 0;
}
‘柒’ C语言编程求以下公式的值 s=1+1/(1*2)+1/(2*3)+...+1/(n*(n+1))答案
结果可以验证 s=1+1/(1*2)+1/(2*3)+...+1/(n*(n+1)) = 1+1-1/n+1
#include "stdafx.h"
#include <iostream>
#include <iomanip>
// s=1+1/(1*2)+1/(2*3)+...+1/(n*(n+1))
#define N 1000
int _tmain(int argc, _TCHAR* argv[])
{
double dResult, dResult1;
int n=1;
dResult=1;
while(n<N) { dResult += 1.0/n/(n+1); n++;}
dResult1 = 1 + 1 - 1.0/(N+1);
std::cout<<std::setprecision(5)<<"直接计算结果="<<dResult<<std::endl;
std::cout<<std::setprecision(5)<<"参考验证结果="<<dResult1<<std::endl;
return 0;
}
运行结果：
直接计算结果=1.999
参考验证结果=1.999
‘捌’ 编写C语言程序，计算下列公式中S的值（n是运行程序时输入的一个正整数）
#include <stdio.h>
int fun1(int);
int fun2(int);
void main(){
int n;
printf("please input an integer:");
scanf("%d",&n);
printf("1+(1+2)+(1+2+3)+......+(1+2+3+.......+n)=%d\n",fun1(n));
printf("1*2-2*3+3*4-4*5+......+(-1)^(n-1)*n*(n+1)=%d\n",fun2(n));
}
int fun1(int n){
int i,s=0,S=0;
for(i=1;i<=n;i++){
s=s+i;
S=S+s;
}
return S;
}
int fun2(int n){
int i,S=0,p=1;
for(i=1;i<=n;i++){
S=S+p*i*(i+1);
p=-p;
}
return S;
}
‘玖’ 这个方程在C语言中怎样编写，求这个方程的程序代码
按照题目要求编写的计算方程的值的C语言程序如下
#include<stdio.h>
int main(){
double x=5,y;
y=5*x*x+4*x-6;
printf("%lf ",y);
return 0;
}
