多项式A除以B
这仍然是一道关于A/B的题，只不过A和B都换成了多项式。你需要计算两个多项式相除的商Q和余R，其中R的阶数必须小于B的阶数。
输入格式：
输入分两行，每行给出一个非零多项式，先给出A，再给出B。每行的格式如下：
N e[1] c[1] ... e[N] c[N]其中N是该多项式非零项的个数，e[i]是第i个非零项的指数，c[i]是第i个非零项的系数。各项按照指数递减的顺序给出，保证所有指数是各不相同的非负整数，所有系数是非零整数，所有整数在整型范围内。
输出格式：
分两行先后输出商和余，输出格式与输入格式相同，输出的系数保留小数点后1位。同行数字间以1个空格分隔，行首尾不得有多余空格。注意：零多项式是一个特殊多项式，对应输出为0 0 0.0。但非零多项式不能输出零系数（包括舍入后为0.0）的项。在样例中，余多项式其实有常数项-1/27，但因其舍入后为0.0，故不输出。
输入样例：
4 4 1 2 -3 1 -1 0 -1
3 2 3 1 -2 0 1输出样例：
3 2 0.3 1 0.2 0 -1.0
1 1 -3.1 ans_xishu[i] = (double) ((int)(ans_xishu[i]*10 + (ans_xishu[i]<0?-0.5:0.5)))/10;#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
struct dxs{
int zhishu;
double xishu;
}c[33000];
bool cmp(dxs d1,dxs d2)
{
return d1.zhishu>d2.zhishu;
}
int main()
{
// freopen("test.in","r",stdin);
// freopen("test.out","w",stdout);
int a_zhishu[33000],b_zhishu[33000],temp_zhishu[33000],shuchu;
double temp_xishu[33000],b_xishu[33000];
double ans_xishu[33000],a_xishu[33000];
int ans_zhishu[33000];
bool temp_biaoji[33000],pd,vo;
int n1,n2,i,num=0,start,tot=0;
cin>>n1;
for (i=1;i<=n1;i++)
scanf("%d%lf",&a_zhishu[i],&a_xishu[i]);
cin>>n2;
for (i=1;i<=n2;i++)
scanf("%d%lf",&b_zhishu[i],&b_xishu[i]);
while (n1!=0 and a_zhishu[1]>=b_zhishu[1])
{
start=1;
num++;
ans_zhishu[num]=a_zhishu[1]-b_zhishu[1];
ans_xishu[num]=a_xishu[1]/b_xishu[1];
for (i=1;i<=n2;i++)
{
temp_zhishu[i]=ans_zhishu[num]+b_zhishu[i];
temp_xishu[i]=ans_xishu[num]*b_xishu[i];
temp_biaoji[i]=0;
}
for (i=1;i<=n1;i++)
{
pd=false;
for (int j=start;j<=n2 and temp_zhishu[j]>=a_zhishu[i];j++)
if (a_zhishu[i]==temp_zhishu[j])
{
pd=true;
if (a_xishu[i]!=temp_xishu[j])
{
c[++tot].zhishu=a_zhishu[i];
c[tot].xishu=a_xishu[i]-temp_xishu[j];
}
temp_biaoji[j]=1;
start++;
break;
}
if (pd==false)
{
c[++tot].zhishu=a_zhishu[i];
c[tot].xishu=a_xishu[i];
}
}
for (i=1;i<=n2;i++)
if (temp_biaoji[i]==0)
{
c[++tot].zhishu=temp_zhishu[i];
c[tot].xishu=(-1)*temp_xishu[i];
}
sort(c+1,c+tot+1,cmp);
for (int i=1;i<=tot;i++)
{
a_xishu[i]=c[i].xishu;
a_zhishu[i]=c[i].zhishu;
}
n1=tot;
tot=0;
}
for(int i=1;i<=num;i++)
ans_xishu[i] = (double) ((int)(ans_xishu[i]*10 + (ans_xishu[i]<0?-0.5:0.5)))/10;
for(int i=1;i<=n1;i++)
a_xishu[i] = (double)((int)(a_xishu[i]*10 + (a_xishu[i]<0?-0.5:0.5)))/10;
shuchu=num;
for (i=1;i<=num;i++)
if (ans_xishu[i]==0)
shuchu--;
if (shuchu==0)
cout<<"0 0 0.0"<<endl;
else
{
vo=true;
printf("%d ",shuchu);
for (i=1;i<=num;i++)
if (ans_xishu[i]!=0)
if (vo==true)
{
printf("%d %.1lf",ans_zhishu[i],ans_xishu[i]);
vo=false;
}
else
printf(" %d %.1lf",ans_zhishu[i],ans_xishu[i]);
printf("\n");
}
shuchu=n1;
for (i=1;i<=n1;i++)
if (a_xishu[i]==0)
shuchu--;
if (shuchu==0)
cout<<"0 0 0.0";
else
{
printf("%d ",shuchu);
vo=true;
for (i=1;i<=n1;i++)
if (a_xishu[i]!=0)
if (vo==true)
{
printf("%d %.1lf",a_zhishu[i],a_xishu[i]);
vo=false;
}
else
printf(" %d %.1lf",a_zhishu[i],a_xishu[i]);
}
return 0;
}
#include <cstdio>
#include <cmath>
using namespace std;
int nonNegativeNum(double c[], int start) {
int cnt = 0;
for (int i = start; i >= 0; i--)
if (abs(c[i]) + 0.05 >= 0.1) cnt++;
return cnt;
}
void printPoly(double c[], int start) {
printf("%d", nonNegativeNum(c, start));
if (nonNegativeNum(c, start) == 0) printf(" 0 0.0");
for (int i = start; i >= 0; i--)
if (abs(c[i]) + 0.05 >= 0.1)
printf(" %d %.1f", i, c[i]);
}
double c1[3000], c2[3000], c3[3000];
int main() {
int m = 0, n = 0, t = 0, max1 = -1, max2= -1;
scanf("%d", &m);
for (int i = 0; i < m; i++) {
scanf("%d", &t);
max1 = max1 > t ? max1 : t;
scanf("%lf", &c1[t]);
}
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &t);
max2 = max2 > t ? max2 : t;
scanf("%lf", &c2[t]);
}
int t1 = max1, t2 = max2;
while (t1 >= t2) {
double c = c1[t1] / c2[t2];
c3[t1 - t2] = c;
for (int i = t1, j = t2; j >= 0; j--, i--) c1[i] -= c2[j] * c;
while (abs(c1[t1]) < 0.000001) t1--;
}
printPoly(c3, max1 - max2);
printf("\n");
printPoly(c1, t1);
return 0;
}