H. GSS and Simple Math Problem--“今日头条杯”首届湖北省大学程序设计竞赛（网络同步赛）

这套题的github地址（里面包含了数据，题解，现场排名）：点此

题目描述

Given n positive integers $a_1, a_2, \ldots, a_n$ , your task is to calculate the product of these integers, The answer is less than $10^{10^5}$

输入描述:

The first line of input is an integer n, the i-th of the following n lines contains the integer

输出描述:

Output one line with the answer

$\prod a_i$

输入例子:

输出例子:

-->

示例1

输入

输出

360360
题目意思：就是给你n个整数，求相乘的大小。

这题python和java很好过，因为py和java有大数，c++就比较难受了

n=int(input())

ans=1

for i in range(n):

    b=int(input())

    ans=ans*b

print(ans)

c++的后来在补

fft代码：超时了，回头再优化

@@ -, +, @@

#include<cmath>

#include<cstdio>

#include<vector>

#include<queue>

#include<cstring>

#include<iomanip>

#include<stdlib.h>

#include<iostream>

#include<algorithm>

#define ll long long

#define inf 1000000000

#define mod 1000000007

#define N 350000

#define fo(i,a,b) for(i=a;i<=b;i++)

#define fd(i,a,b) for(i=a;i>=b;i--)

using namespace std;

const double pi = 3.141592653;

char s1[N>>],s2[N>>];

double rea[N],ina[N],reb[N],inb[N],ret[N],intt[N];

int i,len1,len2,lent,lenres,len;

int res[N>>];

void FFT(double *reA,double *inA,int n,int flag)

{

    if (n == ) return;

    int k,u,i;

    double reWm = cos(*pi/n) , inWm = sin(*pi/n);//��ԭ��

    if (flag) inWm = -inWm;

    double reW = 1.0 , inW = 0.0;

    for (k = ,u = ;k < n; k += ,u++)//�������ż����ֿ�

        {ret[u] = reA[k]; intt[u] = inA[k];}

    for (k = ;k < n; k += )

        {reA[k/] = reA[k]; inA[k/] = inA[k];}

    for (k = u,i = ;k < n && i < u; k++,i++)

        {reA[k] = ret[i]; inA[k] = intt[i];}

    FFT(reA,inA,n/,flag); FFT(reA+n/,inA+n/,n/,flag);

    fo(k,,n/-)//�ϲ�

        {

            int tag = n /  + k;

            double reT = reW * reA[tag] - inW * inA[tag];

            double inT = reW * inA[tag] + inW * reA[tag];

            double reU = reA[k] , inU = inA[k];

            reA[k] = reU + reT; inA[k] = inU + inT;

            reA[tag] = reU - reT; inA[tag] = inU - inT;

            double reWt = reW * reWm - inW * inWm;

            double inWt = reW * inWm + inW * reWm;

            reW = reWt; inW = inWt;

        }

}

void mul()

{

        memset(res,  , sizeof(res));

        memset(rea,  , sizeof(rea));

        memset(ina,  , sizeof(ina));

        //memset(reb, 0 , sizeof(reb));

       // memset(inb, 0 , sizeof(inb));

        len1 = strlen(s1); len2 = strlen(s2);

        lent = (len1 > len2 ? len1 : len2);

        len = ;

        while (len < lent) len <<= ;

        len <<= ;

        fo(i,,len-)

        {

            if (i < len1) rea[i] = (double) s1[len1-i-] - '';

            if (i < len2) reb[i] = (double) s2[len2-i-] - '';

            ina[i] = inb[i] = 0.0;

        }

    FFT(rea,ina,len,); FFT(reb,inb,len,);//���a��b�ĵ�ֵ��ʾ��

    fo(i,,len-)//���c�ĵ�ֵ��ʾ��

        {

            //printf("%.5lf %.5lf\n",rea[i],ina[i]);

            double rec = rea[i] * reb[i] - ina[i] * inb[i];

            double inc = rea[i] * inb[i] + ina[i] * reb[i];

            rea[i] = rec; ina[i] = inc;

        }

    FFT(rea,ina,len,);//���c��ϵ����ʾ��

    fo(i,,len-) {rea[i] /= len; ina[i] /= len;}

    fo(i,,len-) res[i] = (int)(rea[i] + 0.5);

    fo(i,,len-) res[i+] += res[i] /  , res[i] %= ;

    lenres = len1 + len2 + ;

    while (res[lenres] ==  && lenres > ) lenres--;

    int kk=;

    fd(i,lenres,)

        s1[kk++]=res[i]+'';

}

int main()

{

       freopen("temin.txt","r",stdin);

    freopen("temout.txt","w",stdout);

    int n;

    scanf("%d",&n);

    if(n==)

    {

        scanf("%s",s1);

        printf("%s\n",s1);

    }

    else

    {

        scanf("%s",s1);

        for(int i=;i<n;i++)

        {

            scanf("%s",s2);

            mul();

        }

        printf("%s\n",s1);

    }

    return ;

}

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