9.6模拟赛
作者：互联网
1 #include<bits/stdc++.h>
2 #define ll long long
3 using namespace std;
4 const ll N=51;
5 ll n,m,a[N],lj[N],ans,sum,len,tmp;
6
7 inline ll re_ad() {
8 char ch=getchar(); ll x=0,f=1;
9 while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
10 while('0'<=ch && ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
11 return x*f;
12 }
13
14 inline ll gcd(ll x,ll y) { return !y?x:gcd(y,x%y); }
15
16 void dfs(ll d) {
17 // printf("%lld\n",d);
18 if(d>=n+1) {
19 // printf("%lld %lld\n",len,sum);
20 tmp=((len&1)?-1:1)*sum;
21 ans+=m/tmp;
22 return ;
23 }
24 dfs(d+1);
25 if(a[d]>m) return ;
26 ll G=gcd(sum,a[d]);
27 // printf("gcd: %lld %lld %lld\n",sum,a[d],G);
28 if(sum/G*a[d]>m) return ;
29 sum=sum/G*a[d],lj[d]=1,++len;
30 dfs(d+1);
31 sum=sum/a[d]*G,lj[d]=0,--len;
32 }
33
34 int main()
35 {
36 m=re_ad(),n=re_ad();
37 for(ll i=1;i<=n;++i) a[i]=re_ad();
38 if(n==0) { printf("%lld\n",m); return 0; }
39 sum=1,dfs(1);
40 printf("%lld\n",ans);
41 // system("pause");
42 return 0;
43 }
1 #include<bits/stdc++.h>
2 #define ll long long
3 using namespace std;
4 const ll N=4e5+11,Tree_Sz=1e6+6e5+11;
5 ll n,nn,k,a[N],b[N],seg[N];
6
7 inline ll re_ad() {
8 char ch=getchar(); ll x=0,f=1;
9 while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
10 while('0'<=ch && ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
11 return x*f;
12 }
13
14 namespace ST {
15 ll val[2][Tree_Sz];
16 inline void Init() {
17 for(int i=0;i<=(n<<3)+(n<<1);++i) val[0][i]=val[1][i]=0;
18 }
19 inline void Pushup(ll d,ll op) {
20 val[op][d]=val[op][d<<1]+val[op][d<<1|1];
21 }
22 void modify(ll d,ll l,ll r,ll x,ll z,ll op) {
23 if(l==r) {
24 val[op][d]+=z;
25 return ;
26 }
27 ll mid=(l+r)>>1;
28 if(x<=mid) modify(d<<1,l,mid,x,z,op);
29 else modify(d<<1|1,mid+1,r,x,z,op);
30 Pushup(d,op);
31 }
32 inline void Modify(ll x,ll z,ll op) {
33 modify(1,1,n*2,x+n,z,op);
34 }
35 ll query(ll d,ll l,ll r,ll x,ll y,ll op) {
36 if(x<=l && r<=y) {
37 return val[op][d];
38 }
39 ll mid=(l+r)>>1,res=0;
40 if(x<=mid) res+=query(d<<1,l,mid,x,y,op);
41 if(y>=mid+1) res+=query(d<<1|1,mid+1,r,x,y,op);
42 return res;
43 }
44 inline ll Query(ll x,ll op) {
45 return query(1,1,n*2,x+n,n*2,op);
46 }
47 }
48
49 inline bool check(ll d) {
50 ST::Init();
51 ll res=0,lj=0;
52 for(ll i=1;i<=n;++i) seg[i]=(a[i]>d)?1:-1;
53 ST::Modify(seg[1],1,0);
54 res+=(seg[1]==1)?1:0;
55 for(ll i=2,j=1;i<=n;++i,j^=1) {
56 lj-=seg[i];
57 ST::Modify(seg[i]+lj,1,j);
58 // printf("awa\n");
59 res+=ST::Query(lj+1,j);
60 }
61 // printf("check: %lld %lld\n",d,res);
62 if(res<k) return true;
63 return false;
64 }
65
66 inline ll Binary_Find() {
67 ll l=1,r=n,ans=0;
68 while(l<=r) {
69 ll mid=(l+r)>>1;
70 // printf("mid: %lld %lld %lld\n",l,r,mid);
71 if(check(b[mid])) r=mid-1,ans=mid;
72 else l=mid+1;
73 }
74 return b[ans];
75 }
76
77 int main()
78 {
79 n=re_ad(),k=re_ad();
80 for(ll i=1;i<=n;++i) b[i]=a[i]=re_ad();
81 sort(b+1,b+n+1);
82 for(ll i=n;i>=1;i-=2) nn+=i;
83 // printf("%lld\n",nn);
84 printf("%lld\n",Binary_Find());
85 // for(ll i=1;i<=nn;++i) printf("%lld ",check(i)); puts("");
86 // system("pause");
87 return 0;
88 }
1 #include<bits/stdc++.h>
2 #define ll long long
3 using namespace std;
4 const ll N=2e6+11,mod=1e9+7;
5 ll n,a[N],A,B,C,b[N],num,c[N];
6
7 namespace BIT1 {
8 ll t[N];
9 inline ll lowbit(ll d) { return d&(-d); }
10 inline void insert(ll d,ll x) { for(;d<=num+1;d+=lowbit(d)) t[d]+=x; }
11 inline ll query(ll d) { ll res=0; for(;d>0;d-=lowbit(d)) res+=t[d]; return res; }
12 }
13
14 namespace BIT2 {
15 ll t[N];
16 inline ll lowbit(ll d) { return d&(-d); }
17 inline void insert(ll d,ll x) { for(;d<=num+1;d+=lowbit(d)) t[d]+=x; }
18 inline ll query(ll d) { ll res=0; for(;d>0;d-=lowbit(d)) res+=t[d]; return res; }
19 }
20
21 int main()
22 {
23 scanf("%lld%lld%lld%lld%lld",&n,&a[1],&A,&B,&C);
24 b[1]=a[1];
25 for(ll i=2;i<=n;++i) b[i]=a[i]=((a[i-1]*A)%C+B)%C;
26 // printf("a: "); for(ll i=1;i<=n;++i) printf("%lld ",a[i]); puts("");
27 sort(b+1,b+n+1);
28 num=unique(b+1,b+n+1)-b-1;
29 // printf("b: "); for(ll i=1;i<=n;++i) printf("%lld ",b[i]); puts("");
30 for(ll i=1;i<=n;++i) c[i]=lower_bound(b+1,b+num+1,a[i])-b;
31 // printf("c: "); for(ll i=1;i<=n;++i) printf("%lld ",c[i]); puts("");
32 ll ans=0;
33 for(ll i=1;i<=n;++i) {
34 ll t1=BIT1::query(c[i]-1+1)%mod;
35 // printf("%lld ",t1);
36 (t1*=a[i]*(n-i+1)%mod)%=mod;
37 // printf("%lld ",t1);
38 ll t2=(BIT2::query(num+1)-BIT2::query(c[i]+1))%mod;
39 // printf("%lld ",t2);
40 (t2*=(n-i+1))%=mod;
41 // printf("%lld ",t2);
42 (ans+=t1)%=mod;
43 // printf("%lld ",ans);
44 (ans+=t2)%=mod;
45 // printf("%lld\n",ans);
46 BIT1::insert(c[i]+1,i); //printf("insert1: %lld %lld\n",a[i]+1,i);
47 BIT2::insert(c[i]+1,(i*a[i])%mod);
48 // printf("test: %lld %lld %lld\n",BIT1::query(4),BIT1::query(5),BIT1::query(6));
49 }
50 for(ll i=1;i<=n;++i) (ans+=(i*(n-i+1)%mod)*a[i]%mod)%=mod;
51 printf("%lld\n",ans);
52 // if(n>1e5) printf("%lld\n",n);
53 return 0;
54 }
标签：ch,return,res,sum,9.6,ll,模拟,lld 来源： https://www.cnblogs.com/shangguanshufang/p/9_6_mo_ni_sai.html