2018icpc南京现场赛-G Pyramid

#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);
    }
}

https://www.cnblogs.com/wrjlinkkkkkk/p/10041144.html

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。

标签：24,Pyramid,现场,long,int,2018icpc,ans,include,mod
来源： https://www.cnblogs.com/2462478392Lee/p/11552228.html