PAT 1002. A+B for Polynomials (25)
2015-04-14 20:47
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这个道题和1009相似,但是因为忘记结果可能小于0,就悲剧的交了好几次,依旧比较水,但总是不能想的全面,反思~
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int k1,k2;
while(cin>>k1)
{
double sum[1002];
int n,num=0;
double a;
for(int i=1000;i>=0;i--)
{
sum[i]=0.0;
}
for(int i=0;i<k1;i++)
{
cin>>n>>a;
sum
=a;
}
cin>>k2;
for(int i=0;i<k2;i++)
{
cin>>n>>a;
sum
+=a;
}
for(int i=0;i<=1000;i++)
{
if(fabs(sum[i])>1e-6)
{
num++;
}
}
printf("%d",num);
for(int i=1000;i>=0;i--)
{
if(fabs(sum[i])>1e-6)
printf(" %d %.1lf",i,sum[i]);
}
printf("\n");
}
return 0;
}
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int k1,k2;
while(cin>>k1)
{
double sum[1002];
int n,num=0;
double a;
for(int i=1000;i>=0;i--)
{
sum[i]=0.0;
}
for(int i=0;i<k1;i++)
{
cin>>n>>a;
sum
=a;
}
cin>>k2;
for(int i=0;i<k2;i++)
{
cin>>n>>a;
sum
+=a;
}
for(int i=0;i<=1000;i++)
{
if(fabs(sum[i])>1e-6)
{
num++;
}
}
printf("%d",num);
for(int i=1000;i>=0;i--)
{
if(fabs(sum[i])>1e-6)
printf(" %d %.1lf",i,sum[i]);
}
printf("\n");
}
return 0;
}
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