[排列组合 + 分段打表] 51nod 算法马拉松25 A. 二分答案

题意

题解

分析一下可得，如果二分最后要停在k位置，需要有一些位置上的值满足与m的一些大小关系，而且关系都是确定的。
直接模拟一趟就可以得到：有num1个位置需要满足 a[i]<=m ，有num2的位置需要满足 a[i]>m ，其他位置显然可以随便乱放，因为根本就不会访问到它们。
这样就能直接写出答案的表达式了：

Pnum1m∗Pnum2n−m∗(n−num1−num2)!

问题来了，如何求大数(1e+9)阶乘模1e9+7的值呢？
由于数字范围和模数范围都很大，这个东西不好直接求，所以这里需要用到一些奇技淫巧，分段打表登场。
大概就是均匀的每隔一段打一个，然后算阶乘时找到表中最相邻的，然后暴力一步一步推即可。
然后复杂度就被强行除了一个数。又解锁了新姿势。

表不大，代码就直接贴了。

#include
#include
using namespace std;
typedef long long LL;
const LL MOD=1000000007,blk=5000000;
const int lst[201]={1,974067448,682498929,598816162,491101308,586350670,76479948,463847391,723816384,172827403,67347853,407719831,27368307,606322308,625544428,1669644,199888908,534491822,888050723,884343068,927880474,112249297,281863274,770511792,661224977,935080803,623534362,797848181,970055531,232253360,261384175,659224434,195888993,509096183,66404266,785113347,547665832,996122673,109838563,34538816,933245637,911398531,724691727,114985663,368925948,464456846,268838846,938269070,136026497,564758715,112390913,167240465,135498044,889410460,217544623,996097969,419363534,607730875,500780548,651081062,668123525,563432246,128487469,318951960,30977140,93940075,522049725,559947225,309058615,624577416,386027524,716986954,189239124,179518046,148528617,529726489,940567523,473718967,917084264,105419548,429277690,937409008,996164327,418537348,358655417,775305697,568392357,392838702,780072518,596737944,462639908,843321396,275105629,432030917,909210595,321685608,99199382,5003231,703397904,863250534,733333339,180898306,97830135,864869025,608823837,46819124,256141983,491415383,141827977,688809790,696628828,457469634,637939935,502297454,811575797,48053248,848924691,26011548,131772368,271198437,724464507,510650790,272814771,564188856,326159309,44135644,456152084,816929577,903466878,218107212,92255682,196345098,769795511,407518072,373745190,147050765,606241871,666493603,825871994,561011609,957939114,765215899,435887178,201339230,852304035,469928208,663307737,414236650,375297772,211487466,217598709,676526196,624148346,439411911,671734977,644050694,624500515,715264908,748510389,819801784,203191898,340030191,423951674,331910086,629786193,341080135,672850561,684748812,814362881,60625018,823845496,175638827,116667533,163928347,256473217,393556719,627655552,687265514,245795606,36292292,586445753,953634340,172114298,260466949,193781724,756154604,778983779,49031023,83868974,954913,315103615,346966053,965785236,321900901,492741665,532702135,377329025,847645126,847549272,3258987,698611116};
LL n,m,K,num1,num2;
LL Fact(LL x){
    if(x<0) return 0;
    LL res=lst[x/blk];
    for(LL i=(x/blk)*blk+1;i<=x;i++) res=(res*i)%MOD;
    return res;
}
LL power(LL a,LL b){
    if(b==0) return 1;
    if(b&1) return (power(a,b-1)*a)%MOD;
    LL t=power(a,b/2); return (t*t)%MOD;
}
LL P(LL a,LL b){
    if(a<0||b<0||areturn 0;
    return (Fact(a)*power(Fact(a-b),MOD-2))%MOD;
}
int main(){
    freopen("51nod25A.in","r",stdin);
    freopen("51nod25A.out","w",stdout);
    scanf("%lld%lld%lld",&n,&m,&K);
    LL L=1,R=n,mid=(L+R)>>1;
    while(L<=R){
        if (mid<=K) L=mid+1, num1++; // a[mid]<=m
               else R=mid-1, num2++; // a[mid]>m
        mid=(L+R)>>1;
    }
    printf("%lld\n",(P(m,num1)*P(n-m,num2)%MOD)*Fact(n-num1-num2)%MOD);
    return 0;
}