51NOD1799 二分答案 【組合+分塊打表】

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顯然確定了2分的結果下標,模擬一下即可得到途經的所有mid
記錄a[mid]>m 和 a[mid]<=m的mid數量分別爲larger,samller
ans=Clargern−m∗(larger!)∗Csmallerm∗(smaller!)∗[(n−larger∗smaller)!]
要求解1e9!,分塊打表即可

#include<stdio.h>
#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define MEM(a,x) memset(a,x,sizeof(a))
#define lowbit(x) ((x)&-(x))

using namespace std;

const int inf=1e9+7;
const int INF = inf;

pll arr[]={
    {1,1},{10000000,682498929},{20000000,491101308},{30000000,76479948},{40000000,723816384},{50000000,67347853},{60000000,27368307},{70000000,625544428},{80000000,199888908},{90000000,888050723},{100000000,927880474},{110000000,281863274},{120000000,661224977},{130000000,623534362},{140000000,970055531},{150000000,261384175},{160000000,195888993},{170000000,66404266},{180000000,547665832},{190000000,109838563},{200000000,933245637},{210000000,724691727},{220000000,368925948},{230000000,268838846},{240000000,136026497},{250000000,112390913},{260000000,135498044},{270000000,217544623},{280000000,419363534},{290000000,500780548},{300000000,668123525},{310000000,128487469},{320000000,30977140},{330000000,522049725},{340000000,309058615},{350000000,386027524},{360000000,189239124},{370000000,148528617},{380000000,940567523},{390000000,917084264},{400000000,429277690},{410000000,996164327},{420000000,358655417},{430000000,568392357},{440000000,780072518},{450000000,462639908},{460000000,275105629},{470000000,909210595},{480000000,99199382},{490000000,703397904},{500000000,733333339},{510000000,97830135},{520000000,608823837},{530000000,256141983},{540000000,141827977},{550000000,696628828},{560000000,637939935},{570000000,811575797},{580000000,848924691},{590000000,131772368},{600000000,724464507},{610000000,272814771},{620000000,326159309},{630000000,456152084},{640000000,903466878},{650000000,92255682},{660000000,769795511},{670000000,373745190},{680000000,606241871},{690000000,825871994},{700000000,957939114},{710000000,435887178},{720000000,852304035},{730000000,663307737},{740000000,375297772},{750000000,217598709},{760000000,624148346},{770000000,671734977},{780000000,624500515},{790000000,748510389},{800000000,203191898},{810000000,423951674},{820000000,629786193},{830000000,672850561},{840000000,814362881},{850000000,823845496},{860000000,116667533},{870000000,256473217},{880000000,627655552},{890000000,245795606},{900000000,586445753},{910000000,172114298},{920000000,193781724},{930000000,778983779},{940000000,83868974},{950000000,315103615},{960000000,965785236},{970000000,492741665},{980000000,377329025},{990000000,847549272},{1000000000,698611116}
};

const int N = 50 + 5;
int larger,smaller;
void binS(int n,int k){
    larger=smaller=0;
    int l=1,r=n;
    int mid=(l+r)/2;
    while (l<=r)
    {
        if (mid<=k){
            l=mid+1;
            ++smaller;
        }
        else{
            r=mid-1;
            ++larger;
        }
        mid=(l+r)/2;
    }
}

ll getFact(ll n){
    if(n<=0){
        return 1;
    }
    int m=sizeof(arr)/sizeof(*arr);
    pll*it=lower_bound(arr,arr+m,make_pair(n,(ll)0));
    ll ans;
    if(it->first>n){
        --it;
    }
    ans=it->second;
    while(n>it->first){
        ans=(ans*n)%inf;
        --n;
    }
    return ans;
}

ll slove(int n,int m){
    ll ans=1;
    for(int i=0;i<larger;++i){
        ans=(ans*(n-m-i))%inf;
    }
    for(int i=0;i<smaller;++i){
        ans=(ans*(m-i))%inf;
    }
    return (ans*getFact(n-smaller-larger))%inf;
}

int main(){
    //freopen("/home/lu/code/r.txt","r",stdin);
    //freopen("/home/lu/code/w.txt","w",stdout);
    int n,m,k;
    while(~scanf("%d%d%d",&n,&m,&k)){
        binS(n,k);
        printf("%lld\n",slove(n,m));
    }
    return 0;
}