江西财经大学第一届程序设计竞赛 H题 求大数的阶乘

链接：https://www.nowcoder.com/acm/contest/115/H

来源：牛客网

晚上，小P喜欢在寝室里一个个静静的学习或者思考，享受自由自在的单身生活。

他总是能从所学的知识散发出奇妙的思维。

今天他想到了一个简单的阶乘问题，

0！= 1

1！= 1

2！= 1 * 2 = 2

3！= 1 * 2 * 3 = 6

4！= 1 * 2 * 3 *4 = 24

5！= 1 * 2 * 3 *4 * 5 = 120

。

。

如果 n=1000000000，那么n的阶乘会是多少呢，小P当然知道啦，那么你知道吗？

输入描述:

第一行输入一个整数T（表示样例个数）
接下来T组样例
每组样例一行，输入一个整数N（0<=N<=1000000000）

输出描述:

输出T行
每一行输出N的阶乘 N！（由于这个数比较大，所以只要输出其对1000000007取膜的结果即可）

示例1

输入

2
0
1000000000

输出

1
698611116


因为时间限制，所以这题主要是先打表，在n小于10000000时，电脑求阶乘所需的时间比较小，所以
我们先把1~1000000000之间的数分成100个区间,每个区间有10000000个数字，例如[1,10000000)
[10000000,20000000),[20000000,30000000)......
然后我们现在电脑上把每个区间的第一个数的值求出来，例如10000000！%mod=682498929，20000000!%mod=491101308...
把这些值先存进数组a中，然后每次输入n时就判断在哪个区间，把需要计算的次数缩小在10000000次以下，然后暴力求解。

先把每个区间的第一个值算出来：

#include<stdio.h>
#define ll long long
#define mod 1000000007
ll ans;
int main()
{
    ans=1;
    for(ll i=1;i<=1000000000;i++)
    {
        ans=(ans*i)%mod;
        if(i%10000000==0)
        printf("%lld,",ans);
    }
    return 0;
}

682498929,491101308,76479948,723816384,67347853,27368307,625544428,
199888908,888050723,927880474,281863274,661224977,623534362,970055531,261384175,
195888993,66404266,547665832,109838563,933245637,724691727,368925948,268838846,
136026497,112390913,135498044,217544623,419363534,500780548,668123525,128487469,
30977140,522049725,309058615,386027524,189239124,148528617,940567523,917084264,
429277690,996164327,358655417,568392357,780072518,462639908,275105629,909210595,
99199382,703397904,733333339,97830135,608823837,256141983,141827977,696628828,
637939935,811575797,848924691,131772368,724464507,272814771,326159309,456152084,
903466878,92255682,769795511,373745190,606241871,825871994,957939114,435887178,
852304035,663307737,375297772,217598709,624148346,671734977,624500515,748510389,
203191898,423951674,629786193,672850561,814362881,823845496,116667533,256473217,
627655552,245795606,586445753,172114298,193781724,778983779,83868974,315103615,
965785236,492741665,377329025,847549272,698611116

得到上面的数据

最后直接打表计算就可以了

#include<stdio.h>
#include<string.h>
#define ll long long
const ll mod=1e9+7;
ll n,t;
int a[150]={1,682498929,491101308,76479948,723816384,67347853,27368307,625544428,
199888908,888050723,927880474,281863274,661224977,623534362,970055531,261384175,
195888993,66404266,547665832,109838563,933245637,724691727,368925948,268838846,
136026497,112390913,135498044,217544623,419363534,500780548,668123525,128487469,
30977140,522049725,309058615,386027524,189239124,148528617,940567523,917084264,
429277690,996164327,358655417,568392357,780072518,462639908,275105629,909210595,
99199382,703397904,733333339,97830135,608823837,256141983,141827977,696628828,
637939935,811575797,848924691,131772368,724464507,272814771,326159309,456152084,
903466878,92255682,769795511,373745190,606241871,825871994,957939114,435887178,
852304035,663307737,375297772,217598709,624148346,671734977,624500515,748510389,
203191898,423951674,629786193,672850561,814362881,823845496,116667533,256473217,
627655552,245795606,586445753,172114298,193781724,778983779,83868974,315103615,
965785236,492741665,377329025,847549272,698611116};
int main()
{
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld",&n);
        int k=n/10000000;
        ll ans=a[k];
        for(ll i=k*10000000+1;i<=n;i++)
        {
            ans=(ans*i)%mod;
        }
        printf("%lld
",ans);
    }
    return 0;
}