CodeForces 75C Modified GCD [二分+数论]

先求出a和b的最大公约数，找出其所有的因数——sqrt(n)的复杂度，涨姿势了。

然后就是判断所有的因数有没有落在low,high区间里面了——二分即可（upper_bound）

C++版本：

[cpp] #include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
vector x;
int low, high, a, b, n, m, ans;

int main() {
scanf("%d%d", &a, &b);
a = __gcd(a, b);
b = sqrt(a);
x.clear();
for (int i=1; i<=b; i++)
if (a % i == 0) {
x.push_back(i);
x.push_back(a/i);
}
sort(x.begin(), x.end());
scanf("%d", &n);
for (int i=0; i scanf("%d%d", &low, &high);
m = upper_bound(x.begin(), x.end(), high) - x.begin() - 1;
ans = x[m];
if (low > ans) puts("-1");
else printf("%d\n", ans);
}

return 0;
}

#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
vector x;
int low, high, a, b, n, m, ans;

int main() {
scanf("%d%d", &a, &b);
a = __gcd(a, b);
b = sqrt(a);
x.clear();
for (int i=1; i<=b; i++)
if (a % i == 0) {
x.push_back(i);
x.push_back(a/i);
}
sort(x.begin(), x.end());
scanf("%d", &n);
for (int i=0; i scanf("%d%d", &low, &high);
m = upper_bound(x.begin(), x.end(), high) - x.begin() - 1;
ans = x[m];
if (low > ans) puts("-1");
else printf("%d\n", ans);
}

return 0;
}

Python版本：

[python] from fractions import gcd
from bisect import bisect_right as br
g = gcd(*map(int, raw_input().split()))
i = 1
r = []
while i*i <= g:
if g % i == 0:
r.append(i)
r.append(g/i)
i += 1
r = sorted(r)

for i in xrange(input()):
l, h = map(int, raw_input().split())
m = r[br(r, h)-1]
print -1 if m < l else m

from fractions import gcd
from bisect import bisect_right as br
g = gcd(*map(int, raw_input().split()))
i = 1
r = []
while i*i <= g:
if g % i == 0:
r.append(i)
r.append(g/i)
i += 1
r = sorted(r)

for i in xrange(input()):
l, h = map(int, raw_input().split())
m = r[br(r, h)-1]
print -1 if m < l else m