17-10-11模拟赛

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 #define MN 10000105
 6 #define MP 665000
 7 using namespace std;
 8 bool vis[MN];
 9 int pri[MP];
10 long long n,T;
11 int pos,t,cnt=0;
12 inline int in(){
13     int x=0;bool f=0; char c;
14     for (;(c=getchar())<‘0‘||c>‘9‘;f=c==‘-‘);
15     for (x=c-‘0‘;(c=getchar())>=‘0‘&&c<=‘9‘;x=(x<<3)+(x<<1)+c-‘0‘);
16     return f?-x:x;
17 }
18 inline int getp(int n){
19     cnt=0;pri[0]=1;
20     for (int i=2;i<=n;++i){
21         if (!vis[i]) pri[++cnt]=i;
22         for (int j=1;i*pri[j]<=n&&j<=cnt;++j){
23             vis[i*pri[j]]=1;
24             if (!(i%pri[j])) break;
25         }
26     }
27 }
28 int main()
29 {
30     getp(10000020);scanf("%lld",&T);while (T--){
31         scanf("%lld",&n);if (n<=1) {printf("-1\n");continue;}
32         t=(int)sqrt(n);pos=lower_bound(pri+1,pri+cnt+1,t)-pri;
33         if (pri[pos]>t) --pos;
34         printf("%d\n",pos+2);
35     }return 0;
36 }

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define inf 0x7fffffff
 5 #define MN 1002
 6 using namespace std;
 7 inline int in(){
 8     int x=0;bool f=0; char c;
 9     for (;(c=getchar())<‘0‘||c>‘9‘;f=c==‘-‘);
10     for (x=c-‘0‘;(c=getchar())>=‘0‘&&c<=‘9‘;x=(x<<3)+(x<<1)+c-‘0‘);
11     return f?-x:x;
12 }
13 struct que{
14     int q[MN],id[MN];
15     int head,tail;
16     inline void pushmax(int x,int y){
17         while (head<=tail&&x>q[tail]) --tail;
18         ++tail;q[tail]=x,id[tail]=y;
19     }
20     inline void pushmin(int x,int y){
21         while (head<=tail&&x<q[tail]) --tail;
22         ++tail;q[tail]=x,id[tail]=y;
23     }
24     inline void pop(int x){
25         while (head<=tail&&id[head]<=x) ++head;
26     }
27 }mx[MN],mn[MN],mxc,mnc;
28 int c[MN][MN];
29 int n,m,a,b,cnt=0,res,dif;
30 int main()
31 {
32     freopen("anchorman.in","r",stdin);
33     freopen("anchorman.out","w",stdout);
34     n=in();m=in();a=in();b=in();res=inf,cnt=0;
35     for (int i=1;i<=n;++i)
36     for (int j=1;j<=m;++j) c[i][j]=in();
37     for (int i=1;i<=n;++i)
38     mx[i].head=mn[i].head=1,mx[i].tail=mn[i].tail=0;
39     for (int i=1;i<=n;++i)
40     for (int j=1;j<b;++j)
41     mx[i].pushmax(c[i][j],j),mn[i].pushmin(c[i][j],j);
42     for (int j=b;j<=m;++j){
43         mxc.head=mnc.head=1;mxc.tail=mnc.tail=0;
44         for (int i=1;i<=n;++i){
45             mx[i].pushmax(c[i][j],j);mx[i].pop(j-b);
46             mn[i].pushmin(c[i][j],j);mn[i].pop(j-b);
47             mxc.pushmax(mx[i].q[mx[i].head],i);mxc.pop(i-a);
48             mnc.pushmin(mn[i].q[mn[i].head],i);mnc.pop(i-a);
49             dif=mxc.q[mxc.head]-mnc.q[mnc.head];
50             if (i>=a) if (dif==res) ++cnt;
51             else if (dif<res) res=dif,cnt=1;
52         }
53     }printf("%d %d",res,cnt);return 0;
54 }