hdu 5666 Segment 俄罗斯乘法或者套大数板子
Segment
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Problem Description
Silen August does not like to talk with others.She like to find some interesting problems.
Today she finds an interesting problem.She finds a segment x+y=q.The segment intersect the axis and produce a delta.She links some line between (0,0) and the node on the segment whose coordinate are integers.
Please calculate how many nodes are in the delta and not on the segments,output answer mod P.
Today she finds an interesting problem.She finds a segment x+y=q.The segment intersect the axis and produce a delta.She links some line between (0,0) and the node on the segment whose coordinate are integers.
Please calculate how many nodes are in the delta and not on the segments,output answer mod P.
Input
First line has a number,T,means testcase number.
Then,each line has two integers q,P.
q is a prime number,and 2≤q≤1018,1≤P≤1018,1≤T≤10.
Then,each line has two integers q,P.
q is a prime number,and 2≤q≤1018,1≤P≤1018,1≤T≤10.
Output
Output 1 number to each testcase,answer mod P.
Sample Input
1
2 107
Sample Output
0
Source
题解:答案就是((p-1)*(p-2)/2)%mod;p是10^18以内直接乘就会爆;利用俄罗斯乘法的加法性质得到答案;(现场做只会想到java大数。。果然菜鸟。。。)
俄罗斯乘法:http://baike.baidu.com/link?url=vVo1zdml29g80N-BYvpdm2hNGpYwSnGoJsnAJmook4AJBiYUVL_ort5f7XqFJ0yx6zxB5ha90q6-1LD6HxPIaa
俄罗斯乘法代码:
View Code
View Code
#include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll __int64 #define inf 2000000001 int scan() { int res = 0 , ch ; while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) ) { if( ch == EOF ) return 1 << 30 ; } res = ch - '0' ; while( ( ch = getchar() ) >= '0' && ch <= '9' ) res = res * 10 + ( ch - '0' ) ; return res ; } ll eluosimul(ll x,ll y,ll mod) { ll sum=0; while(x) { if(x&1) { sum+=y; sum%=mod; } x>>=1; y*=2; y%=mod; } return sum; } int main() { ll x,y,z,i,t,m,q; scanf("%I64d",&x); while(x--) { ll ans; scanf("%I64d%I64d",&q,&m); if(q%2) ans=eluosimul(q-2,(q-1)/2,m); else ans=eluosimul(q-1,(q-2)/2,m); printf("%I64d\n",ans); } return 0; }
java:
import java.util.*; import java.math.*; public class Main { public static void main(String[] args) { Scanner cin=new Scanner(System.in); int x; x=cin.nextInt(); while(x!=0) { x--; BigInteger c=new BigInteger("2"); BigInteger e=new BigInteger("1"); BigInteger a=cin.nextBigInteger(); BigInteger d=a.subtract(c); BigInteger f=a.subtract(e); BigInteger b=cin.nextBigInteger(); BigInteger ans=f.multiply(d); ans=ans.divide(c); System.out.println(ans.remainder(b)); } } }
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步