HDU 5666 快速乘

Segment

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 360 Accepted Submission(s): 134


[align=left]Problem Description[/align]

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.

[align=left]Input[/align]

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.

[align=left]Output[/align]

Output 1 number to each testcase,answer mod P.

[align=left]Sample Input[/align]

1

2 107

[align=left]Sample Output[/align]

0

[align=left]Source[/align]
BestCoder Round #80

题意：判断直线x+y=q与坐标轴围成的三角形“内”的整数点的个数 很容易推出公式 （q-1）*（q-2）/2

题解： 直接乘会爆long long
快速乘算法 处理
类比快速幂模拟 http://www.2cto.com/kf/201505/396902.html ll kuai(ll q,ll num,ll mod){
ll ans=0;
ll base=q;
while(num){
if(num%2) ans=(ans+base)%mod;
num/=2;
base=(base+base)%mod;
}
return ans%mod;
}

#include<iostream>
#include<cstring>
#include<cstdio>
#define ll __int64
using namespace std;
int t;
ll q,p;
ll ans1,ans2;
ll re;
ll kuai(ll q,ll num,ll mod){
ll ans=0;
ll base=q;
while(num){
if(num%2) ans=(ans+base)%mod;
num/=2;
base=(base+base)%mod;
}
return ans%mod;
}
int main()
{
while(scanf("%d",&t)!=EOF)
{
for(int i=1;i<=t;i++)
{
scanf("%I64d %I64d",&q,&p);
if(q%2==0)
{
ans1=(q-2)/2;
ans1=ans1%p;
ans2=q-1;
}
else
{
ans1=(q-1)/2;
ans1=ans1%p;
ans2=q-2;
}
printf("%I64d\n",kuai(ans1,ans2,p));
}
}
return 0;
}