链接:https://www.nowcoder.com/acm/contest/115/H
来源:牛客网
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)
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; }