2017省夏令营Day7 【快速幂，筛法，矩阵快速幂，线段树】

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #define Mod 1000000007
 5 using namespace std;
 6 long long a,b,n,ans1,ans2;
 7 long long power(long long x){
 8     long long ret=1,tmp=2;
 9     while(x){
10         if(x&1){ret*=tmp; ret%=Mod;}
11         tmp*=tmp; tmp%=Mod; x/=2;
12     }
13     return ret;
14 }
15 int main()
16 {
17     freopen("apexis.in","r",stdin);
18     freopen("apexis.out","w",stdout);
19     scanf("%lld%lld%lld",&a,&b,&n);
20     if(n==0){
21         printf("%lld %lld",a,b);
22         return 0;
23     }
24     if(a<b) swap(a,b);
25     long long tmp=n>>1;
26     long long t=power(tmp);
27     if(n&1) ans1=t*(a+b)%Mod,ans2=t*(a-b)%Mod;
28     else ans1=t*a%Mod,ans2=t*b%Mod;
29     printf("%lld %lld",ans1,ans2);
30     return 0;
31 }

 1 #include<cstdio>
 2 #include<iostream>
 3 #define Max 200005
 4 using namespace std;
 5 int n,m,ans,cnt,maxn;
 6 int f[Max],num[Max],prime[Max];
 7 void shaifa(){
 8     for(int i=1;i<=Max;i++) f[i]=i;
 9     for(int i=2;i<=Max;i++)
10         if(!prime[i]){
11             f[i]--;
12             for(int j=2*i;j<=Max/2;j+=i){
13                 f[j]=f[j]*(i-1)/i;
14                 prime[j]=1;
15             }
16         }
17 }
18 int main()
19 {
20     freopen("quest.in","r",stdin);
21     freopen("quest.out","w",stdout);
22     shaifa();
23     int T; scanf("%d",&T);
24     while(T--){
25         ans=cnt=maxn=0;
26         scanf("%d%d",&n,&m);
27         for(int i=1;i<=n;i++){
28             int x; scanf("%d",&x);
29             num[x]++; maxn=max(maxn,x);
30         }
31         for(int i=1;i<=maxn;i++){
32             cnt=0;
33             for(int j=i;j<=maxn;j+=i) cnt+=num[j];
34             if(cnt>=m) ans=max(ans,f[i]);
35             num[i]=0;
36         }
37         printf("%d\n",ans);
38     }
39     return 0;
40 }

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #define Max 100001
 5 #define Mod 10007
 6 using namespace std;
 7 int n,m,l,r;
 8 long long K,cnt[Max];
 9 struct mat{long long num[2][2];}a,b,c,ret;
10 mat mult(mat x,mat y){
11     memset(c.num,0,sizeof(c.num));
12     for(int i=0;i<2;i++)
13         for(int j=0;j<2;j++)
14             for(int k=0;k<2;k++)
15                 c.num[i][j]=(c.num[i][j]+x.num[i][k]*y.num[k][j])%Mod;
16     return c;
17 }
18 mat matmod(mat x,long long k){
19     memset(ret.num,0,sizeof(ret.num));
20     for(int i=0;i<2;i++) ret.num[i][i]=1;
21     while(k){
22         if(k&1) ret=mult(ret,x);
23         k>>=1; x=mult(x,x);
24     }
25     return ret;
26 }
27 long long Get(long long k){
28     memset(a.num,0,sizeof(a.num));
29     memset(b.num,0,sizeof(b.num));
30     a.num[0][1]=1; a.num[1][0]=1; a.num[1][1]=1;
31     b.num[0][0]=1; b.num[1][0]=1;
32     a=matmod(a,k-1);
33     a=mult(a,b);
34     return a.num[0][0];
35 }
36 int main()
37 {
38     freopen("rabbit.in","r",stdin);
39     freopen("rabbit.out","w",stdout);
40     scanf("%d%d",&n,&m);
41     for(int i=1;i<=n;i++) cnt[i]=1;
42     for(int i=1;i<=m;i++){
43         long long ans=0;
44         scanf("%d%d%lld",&l,&r,&K);
45         if(K==0){
46             for(int j=l;j<=r;j++)
47                 if(cnt[j]!=0) ans=(ans+Get(cnt[j]))%Mod;
48             printf("%lld\n",ans);
49         }
50         else for(int j=l;j<=r;j++) cnt[j]+=K;
51     }
52     return 0;
53 }