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Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1407 Accepted Submission(s): 581
#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; }
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原文地址:http://www.cnblogs.com/program-ccc/p/5814328.html