#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<vector>
#define ll long long
using namespace std;
double a[20][20],b[20];
int n;
int main()
{
n=5;
b[1]=1.0;
b[2]=5.0;
b[3]=15.0;
b[4]=35.0;
b[5]=70.0;
double c[20][20]={{0,0,0,0,0,0},{0,1,1,1,1,1},{0,16,8,4,2,1},{0,81,27,9,3,1},{0,256,64,16,4,1},{0,625,125,25,5,1}};
for(int i=1;i<=n;i++)
{
for(int j=i;j<=n;j++)
{
if(fabs(c[j][i])>1e-8)
{
for(int k=1;k<=n;k++)
{
swap(c[i][k],c[j][k]);
}
swap(b[i],b[j]);
}
}
for(int j=1;j<=n;j++)
{
if(i==j)
continue;
double rate=c[j][i]/c[i][i];
for(int k=i;k<=n;k++)
c[j][k]-=c[i][k]*rate;
b[j]-=b[i]*rate;
}
}
double m1=b[1]/c[1][1];
double m2=b[2]/c[2][2];
double m3=b[3]/c[3][3];
double m4=b[4]/c[4][4];
ll m;
printf("%lf\n",m1);
while(~scanf("%lld",&m))
{
double ans=m1*pow(m,4)+m2*pow(m,3)+m3*pow(m,2)+m4*m+0.1;
//printf("%lf\n%lf\n%lf\n",m1*pow(m,4),m2*pow(m,3),m3*pow(m,2));
ll d=ans;
printf("%lld\n",d);
}
}


先用高斯消元求系数,分别为1/24,6/24,11/24,1/24,0。

然后求出其24的逆元为41666667。

最后得其代码



#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<vector>
#define ll long long
#define mod 1000000007
using namespace std;
long long n;
long long fpow(long long a,long long p) {
long long base=a%mod;
long long ans=1;
while(p) {
if(p&1)
ans=ans*base%mod;
base=base*base%mod;
p=p>>1;
}
return ans%mod;
}

int main()
{
int t;
scanf("%d",&t);
while(t--) {
scanf("%lld",&n);
long long ans=(fpow(n,4)+fpow(n,3)*6+fpow(n,2)*11+n%mod*6)%mod;
ans=ans*41666667%mod;
printf("%lld\n",ans);
}
}



http://xueshu.baidu.com/s?wd=paperuri%3A%281d6cf25a905f1951b577cf287a82e1a3%29&filter=sc_long_sign&tn=SE_xueshusource_2kduw22v&sc_vurl=http%3A%2F%2Fwenku.baidu.com%2Fview%2Fef155ff1a26925c52dc5bf32.html&ie=utf-8&sc_us=12108961739379565888

解释了为什么是四次函数。