POJ 1845 Sumdiv 【二分 || 逆元】

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 30268		Accepted: 7447

Description

Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 9901).

Input

The only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.

Output

The only line of the output will contain S modulo 9901.

Sample Input

2 3

Sample Output

Hint

2^3 = 8.
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output).

Source

Romania OI 2002

题目概括：

给出 A 和 B ，求 A^B 所有因子之和，答案 mod 9901；

解题思路：

可对 A 先进行素因子分解 A = p1^a1 * p2^a2 * p3^a3 * ...... * pn^an

即 A^B = p1^a1*B * p2^a2*B * p3^a3*B * ...... * pn^an*B

那么 A^B 所有因子之和 = (1 + P1¹ + P1² + P1³ + ... + P1^a1*B) * (1 + P2¹ + P2² + P2³ + ... + P2^a2*B) * ......* (1 + Pn¹ + Pn² + Pn³ + ... + Pn^an*B) ;

对于 (1 + P¹ + P² + P³ + ... + P^a1*B) 我们可以

①二分求和（47 ms)

AC code：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 using namespace std;
 9 const LL MOD = 9901;
10 const int MAXN = 1e4+10;
11 LL p[MAXN];
12 bool isprime[MAXN];
13 int cnt;
14 LL A, B;
15 
16 void init()                               //打表预处理素因子
17 {
18     cnt = 1;
19     memset(isprime, 1, sizeof(isprime));
20     for(LL i = 2; i < MAXN; i++){
21         if(isprime[i]){
22             p[cnt++] = i;
23             for(int j = 1; j < cnt && p[j]*i < MAXN; j++)
24                 isprime[p[j]*i] = false;
25         }
26     }
27 }
28 
29 LL qpow(LL x, LL n)
30 {
31     LL res = 1LL;
32     while(n){
33         if(n&1) res = (res%MOD*x%MOD)%MOD;
34         x = (x%MOD*x%MOD)%MOD;
35         n>>=1LL;
36     }
37     return res;
38 }
39 
40 LL sum(LL x, LL n)
41 {
42     if(n == 0) return 1;
43     LL res = sum(x, (n-1)/2);
44     if(n&1){
45         res = (res + res%MOD*qpow(x, (n+1)/2)%MOD)%MOD;
46     }
47     else{
48         res = (res + res%MOD*qpow(x, (n+1)/2)%MOD)%MOD;
49         res = (res + qpow(x, n))%MOD;
50     }
51     return res;
52 }
53 
54 int main()
55 {
56     LL ans = 1;
57     scanf("%lld %lld", &A, &B);
58     init();
59     //printf("%d
", cnt);
60     for(LL i = 1; p[i]*p[i] <= A && i < cnt; i++){          //素因子分解
61         //printf("%lld
 ", p[i]);
62         if(A%p[i] == 0){
63             LL num = 0;
64             while(A%p[i] == 0){
65                 num++;
66                 A/=p[i];
67             }
68             ans = ((ans%MOD*sum(p[i], num*B)%MOD)+MOD)%MOD;
69         }
70     }
71     if(A > 1){
72        // puts("zjy");
73         ans = (ans%MOD*sum(A, B)%MOD+MOD)%MOD;
74     }
75 
76     printf("%lld
", ans);
77 
78     return 0;
79 }

View Code

②应用等比数列求和公式 ( 0ms )

因为要保证求模时的精度，所以要求逆元。

这里用的方法是一般情况都适用的： A/B mod p = A mod (p*B) / B;

但是考虑乘法会爆int，所以自定义一个二分乘法。

AC code:

 1 // 逆元 + 二分乘法
 2 #include <cstdio>
 3 #include <iostream>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <cstring>
 7 #define INF 0x3f3f3f3f
 8 #define LL long long
 9 using namespace std;
10 const int MAXN = 1e4+10;
11 const LL mod  = 9901;
12 int p[MAXN], cnt;
13 bool isp[MAXN];
14 LL A, B;
15 
16 void prime()
17 {
18     memset(isp, 1, sizeof(isp));
19     isp[0] = isp[1] = false;
20     for(int i = 2; i < MAXN; i++){
21         if(isp[i]){
22             p[cnt++] = i;
23             for(int k = 0; k < cnt && i*p[k] < MAXN; k++){
24                 isp[i*p[k]] = false;
25             }
26         }
27     }
28 }
29 
30 LL mutil(LL x, LL y, LL MOD)
31 {
32     LL res = 0;
33     while(y){
34         if(y&1) res = (res+x)%MOD;
35         x = (x+x)%MOD;
36         y>>=1LL;
37     }
38     return res;
39 }
40 
41 LL q_pow(LL x, LL n, LL MOD)
42 {
43     LL res = 1LL;
44     while(n){
45         if(n&1) res = mutil(res, x, MOD)%MOD;
46         x = mutil(x, x, MOD)%MOD;
47         n>>=1LL;
48     }
49     return res;
50 }
51 
52 int main()
53 {
54     LL num = 0;
55     LL ans = 1;
56     scanf("%lld %lld", &A, &B);
57     prime();
58     for(int i = 0; p[i]*p[i] <= A; i++){
59         if(A%p[i] == 0){
60             num = 0;
61             while(A%p[i] == 0){
62                 A/=p[i];
63                 num++;
64             }
65             LL m = mod*(p[i]-1);
66             ans *= (q_pow(p[i], num*B+1, m) + m-1)/(p[i] - 1);
67             ans = ans%mod;
68             //ans += ((q_pow(p[i], num*B, mod*(p[i]-1))-p[i])/(p[i]-1))%mod;
69         }
70     }
71 
72     if(A != 1){
73 //        ans += ((q_pow(A, B, mod*(A-1))-A)/(A-1))%mod;
74         LL m = mod*(A-1);
75         ans*=(q_pow(A, B+1, m)+m-1)/(A-1);
76         ans = ans%mod;
77     }
78 
79     printf("%lld
", ans);
80 
81     return 0;
82 }

View Code

POJ 1845 Sumdiv 【二分 || 逆元】

Sumdiv

Time Limit: 1000MS

Memory Limit: 30000K

Total Submissions: 30268

Accepted: 7447

题目概括：

解题思路：