首页 > 代码库 > HDU5478(快速幂)

HDU5478(快速幂)

2024-08-11 01:53:27   215人阅读

Can you find it

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1407    Accepted Submission(s): 581


Problem Description
Given a prime number C(1≤C≤2×105), and three integers k1, b1, k2 (1≤k1,k2,b1≤109). Please find all pairs (a, b) which satisfied the equation ak1⋅n+b1+ bk2⋅n−k2+1 = 0 (mod C)(n = 1, 2, 3, ...).
 

 

Input
There are multiple test cases (no more than 30). For each test, a single line contains four integers C, k1, b1, k2.
 

 

Output
First, please output "Case #k: ", k is the number of test case. See sample output for more detail.
Please output all pairs (a, b) in lexicographical order. (1≤a,b<C). If there is not a pair (a, b), please output -1.
 

 

Sample Input
23 1 1 2
 

 

Sample Output
Case #1:1 22
 
思路：枚举a。当n=1时，ak1+b+b=0(mod C)，则b=C-ak1+b(mod C)。再利用n=2，验证b是否正确。
#include <cstdio>using namespace std;typedef long long LL;LL C,k1,b1,k2;LL npow(LL x,LL n,LL mod){    LL res=1;    while(n>0)        {        if(n&1)    res=(res*x)%mod;        x=(x*x)%mod;        n>>=1;    }    return res;}int main(){    int cas=0;    while(scanf("%lld%lld%lld%lld",&C,&k1,&b1,&k2)!=EOF)    {        bool tag=false;        printf("Case #%d:\n",++cas);        for(LL a=1;a<C;a++)        {            LL b=C-npow(a,k1+b1,C);            LL x=npow(a,2*k1+b1,C);            LL y=npow(b,k2+1,C);            if((x+y)%C==0)            {                tag=true;                printf("%lld %lld\n",a,b);            }            }        if(!tag)        {            printf("-1\n");        }    }    return 0;}

 

HDU5478(快速幂)


联系
我们