快速幂

Raising Modulo Numbers http://poj.org/problem?id=1995

快速幂取模

 1 #include<cstdio>
 2 typedef __int64 LL;
 3 LL quickpow(LL a,LL b,LL c){//快速幂求(a^b)%c
 4     LL ret=1%c;
 5     for(;b;a=a*a%c,b>>=1){
 6         if(b&1){
 7             ret=ret*a%c;
 8         }
 9     }
10     return ret;
11 }
12 int main(){
13     int t,n;
14     LL mod,a,b;
15     while(~scanf("%d",&t)){
16         while(t--){
17             scanf("%I64d%d",&mod,&n);
18             LL sum=0;
19             while(n--){
20                 scanf("%I64d%I64d",&a,&b);
21                 sum+=quickpow(a,b,mod);
22                 sum%=mod;
23             }
24             printf("%I64d
",sum);
25         }
26     }
27 }

View Code

showproblem.php?pid=4869

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 typedef __int64 LL;
 5 const int M=100010;
 6 const int mod=1000000009;
 7 LL quickpow(LL a,LL b,LL c){//快速幂求(a^b)%c
 8     LL ret=1%c;
 9     for(;b;a=a*a%c,b>>=1){
10         if(b&1){
11             ret=ret*a%c;
12         }
13     }
14     return ret;
15 }
16 LL C[M];
17 LL INV[M];
18 int main() {
19     for(int i=1; i<M; i++) {
20         INV[i]=quickpow(i,mod-2,mod);
21     }
22     int n,m,a;
23     while(~scanf("%d%d",&n,&m)) {
24         C[0]=1;
25         for(int i=1;i<=m;i++){
26             C[i]=C[i-1]*(m-i+1)%mod*INV[i]%mod;
27         }
28         int L=0,R=0,nl,nr,tmp;
29         for(int i=0;i<n;i++){
30             scanf("%d",&a);
31             tmp=min(m-L,a);
32             nr=L+tmp-(a-tmp);
33             tmp=min(R,a);
34             nl=R-tmp+(a-tmp);
35             if(nl>nr) swap(nl,nr);
36             if(L<=a&&a<=R){
37                 if(L%2==a%2){
38                     nl=0;
39                 }
40                 else{
41                     nl=min(nl,1);
42                 }
43             }
44             if((m-R)<=a&&a<=(m-L)){
45                 if((m-L)%2==a%2){
46                     nr=m;
47                 }
48                 else{
49                     nr=max(nr,m-1);
50                 }
51             }
52             if(L>=a) nl=min(nl,L-a);
53             if(m-R>=a) nr=max(nr,R+a);
54             L=nl;
55             R=nr;
56         }
57         int ans=0;
58         for(int i=L;i<=R;i+=2){
59             ans+=C[i];
60             ans%=mod;
61         }
62         printf("%d
",ans);
63     }
64     return 0;
65 }

View Code

end