c++程序问题分数求和
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c++程序问题分数求和
求以下问题不用数组的办法 急
输入n个分数并对他们求和,用约分之后的最简形式表示.
比如:q/p = x1/y1 + x2/y2 +.+ xn/yn,q/p要求是归约之后的形式.如:5/6已经是最简形式,3/6需要规约为1/2,3/1需要规约成3,10/3就是最简形式.PS:分子和分母都没有为0的情况,也没有出现负数的情况
关于输入 第一行的输入n,代表一共有几个分数需要求和 接下来的n行是分数
关于输出 输出只有一行,即归约后的结果
例子输入 2
1/2
1/3 例子输出 5/6
求以下问题不用数组的办法 急
输入n个分数并对他们求和,用约分之后的最简形式表示.
比如:q/p = x1/y1 + x2/y2 +.+ xn/yn,q/p要求是归约之后的形式.如:5/6已经是最简形式,3/6需要规约为1/2,3/1需要规约成3,10/3就是最简形式.PS:分子和分母都没有为0的情况,也没有出现负数的情况
关于输入 第一行的输入n,代表一共有几个分数需要求和 接下来的n行是分数
关于输出 输出只有一行,即归约后的结果
例子输入 2
1/2
1/3 例子输出 5/6
#include<stdio.h>
unsigned gcd ( unsigned m,unsigned n );
void sum( unsigned *sum_x,unsigned *sum_y, unsigned x, unsigned y);
int main()
{
int n,i;
unsigned sum_x, sum_y,x,y;
scanf("%d",&n);
scanf("%u/%u",&sum_x, &sum_y);
for(i=1;i<n;++i) {
scanf("%u/%u",&x, &y);
sum(&sum_x, &sum_y, x, y);
}
if(sum_y != 1) {
printf("%u/%u\n", sum_x, sum_y);
} else {
printf("%u\n", sum_x);
}
return 0;
}
void sum( unsigned *sum_x,unsigned *sum_y, unsigned x, unsigned y) {
unsigned tmp;
*sum_x = *sum_x * y+ x*(*sum_y);
*sum_y = *sum_y * y;
tmp = gcd(*sum_x, *sum_y);
*sum_x /= tmp;
*sum_y /= tmp;
}
unsigned gcd ( unsigned m,unsigned n )
{
unsigned temp;
if (m<n)
{
temp=m;
m=n;
n=temp;
}
if ( m % n == 0)
{
return n;
}
else
{
return gcd ( n,m % n) ;
}
}
再问: 感谢!不过能换成标准c++么?
再答: C++的输入没法处理这种格式输入:%u/%u,C本身就可以当C++的,不会有问题的
再问: 嗯,谢谢了
再答: C++可改为如下,其它部分跟上面的一样:#include<iostream>
using namespace std;
unsigned gcd ( unsigned m,unsigned n );
void sum( unsigned *sum_x,unsigned *sum_y, unsigned x, unsigned y);
int main()
{
int n,i;
char c;
unsigned sum_x, sum_y,x,y;
cin>>n;
cin>>sum_x>>c>>sum_y;
for(i=1;i<n;++i) {
cin>>x>>c>>y;
sum(&sum_x, &sum_y, x, y);
}
if(sum_y != 1) {
cout<<sum_x<<"/"<<sum_y<<endl;
} else {
cout<<sum_x<<endl;
}
return 0;
}
unsigned gcd ( unsigned m,unsigned n );
void sum( unsigned *sum_x,unsigned *sum_y, unsigned x, unsigned y);
int main()
{
int n,i;
unsigned sum_x, sum_y,x,y;
scanf("%d",&n);
scanf("%u/%u",&sum_x, &sum_y);
for(i=1;i<n;++i) {
scanf("%u/%u",&x, &y);
sum(&sum_x, &sum_y, x, y);
}
if(sum_y != 1) {
printf("%u/%u\n", sum_x, sum_y);
} else {
printf("%u\n", sum_x);
}
return 0;
}
void sum( unsigned *sum_x,unsigned *sum_y, unsigned x, unsigned y) {
unsigned tmp;
*sum_x = *sum_x * y+ x*(*sum_y);
*sum_y = *sum_y * y;
tmp = gcd(*sum_x, *sum_y);
*sum_x /= tmp;
*sum_y /= tmp;
}
unsigned gcd ( unsigned m,unsigned n )
{
unsigned temp;
if (m<n)
{
temp=m;
m=n;
n=temp;
}
if ( m % n == 0)
{
return n;
}
else
{
return gcd ( n,m % n) ;
}
}
再问: 感谢!不过能换成标准c++么?
再答: C++的输入没法处理这种格式输入:%u/%u,C本身就可以当C++的,不会有问题的
再问: 嗯,谢谢了
再答: C++可改为如下,其它部分跟上面的一样:#include<iostream>
using namespace std;
unsigned gcd ( unsigned m,unsigned n );
void sum( unsigned *sum_x,unsigned *sum_y, unsigned x, unsigned y);
int main()
{
int n,i;
char c;
unsigned sum_x, sum_y,x,y;
cin>>n;
cin>>sum_x>>c>>sum_y;
for(i=1;i<n;++i) {
cin>>x>>c>>y;
sum(&sum_x, &sum_y, x, y);
}
if(sum_y != 1) {
cout<<sum_x<<"/"<<sum_y<<endl;
} else {
cout<<sum_x<<endl;
}
return 0;
}