Codeforces Round #188 (Div. 1) / 317A Perfect Pair(数学&优化)
2013-08-16 23:51
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A. Perfect Pair
http://codeforces.com/problemset/problem/317/A
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater
than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard.
It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams
or the %I64dspecifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect
one.
Sample test(s)
input
output
input
output
input
output
注意x,y异号的情况可以优化。
优化后复杂度:O(log m)
完整代码:
http://codeforces.com/problemset/problem/317/A
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater
than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard.
It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams
or the %I64dspecifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect
one.
Sample test(s)
input
1 2 5
output
2
input
-1 4 15
output
4
input
0 -1 5
output
-1
注意x,y异号的情况可以优化。
优化后复杂度:O(log m)
完整代码:
/*30ms,0KB*/ #include<cstdio> #include<cmath> #include<algorithm> using namespace std; int main(void) { __int64 x, y, m, count = 0; scanf("%I64d%I64d%I64d", &x, &y, &m); if (m <= max(x, y)) printf("0"); else { if (x <= 0 && y <= 0) printf("-1"); else { ///x,y异号的情况可以优化一下 if (x < 0 && y > 0){ count = ceil((double) - x / y); x+=count*y; } else if (x > 0 && y < 0){ count = ceil((double) - y / x); y+=count*x; } while (max(x, y) < m) { if (x < y) x += y; else y += x ; ++count; } printf("%I64d", count); } } return 0; }
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