Segment
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 360 Accepted Submission(s): 134
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;
}
ll ans=0;
ll base=q;
while(num){
if(num%2) ans=(ans+base)%mod;
num/=2;
base=(base+base)%mod;
}
return ans%mod;
}
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #define ll __int64 5 using namespace std; 6 int t; 7 ll q,p; 8 ll ans1,ans2; 9 ll re; 10 ll kuai(ll q,ll num,ll mod){ 11 ll ans=0; 12 ll base=q; 13 while(num){ 14 if(num%2) ans=(ans+base)%mod; 15 num/=2; 16 base=(base+base)%mod; 17 } 18 return ans%mod; 19 } 20 int main() 21 { 22 while(scanf("%d",&t)!=EOF) 23 { 24 for(int i=1;i<=t;i++) 25 { 26 scanf("%I64d %I64d",&q,&p); 27 if(q%2==0) 28 { 29 ans1=(q-2)/2; 30 ans1=ans1%p; 31 ans2=q-1; 32 } 33 else 34 { 35 ans1=(q-1)/2; 36 ans1=ans1%p; 37 ans2=q-2; 38 } 39 printf("%I64d ",kuai(ans1,ans2,p)); 40 } 41 } 42 return 0; 43 }