Segment

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1765    Accepted Submission(s): 651 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.   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.   Output      Output 1 number to each testcase,answer mod P.   Sample Input 1 2 107   Sample Output 0   Source BestCoder Round #80   Recommend wange2014   |   We have carefully selected several similar problems for you:   6018  6017  6016  6015  6014  别人说是利用二进制，感觉并不是，其实就是相当于把b分解成n个2相乘，这样每次a*2取模，如果b为奇数，最后再加上一个b,.这样就OK了。 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; __int64 p,q; void solve(__int64 a, __int64 b) { __int64 s=0, base=a; while(b) { if(b&1) { s+=base; s%=p; b--; } else { base*=2; base%=p; b=(b>>1); } } printf("%I64d\n", s); } int main() { int t; scanf("%d", &t); while(t--) { scanf("%I64d%I64d", &q, &p); if(q==2) { printf("0\n"); } else { __int64 x=(q-1)/2%p; __int64 y=(q-2)%p; solve(x, y); } } return 0; }