题目链接：
题目大意：
快速求：
[sum_{i=a}^{b}sum_{j=c}^{d}left[operatorname{gcd}(i,j)==d
ight]
]
正文：
这道题和 [POI2007]ZAP-Queries 思路一样，先化简再整除分块。但是这题不能直接化，先考虑 (a=1,c=1) 的情况，用二维前缀和的思想得到答案。
代码：
ll n, m, n1, m1, d, t;
ll pri[N], miu[N], cnt, sum[N];
bool vis[N];
void prework()
{
miu[1] = 1;
for (int i = 2; i <= N - 10; i++)
{
if(!vis[i]) {pri[++cnt] = i, miu[i] = -1;}
for (int j = 1; j <= cnt && pri[j] * i <= N - 10; j++)
{
vis[pri[j] * i] = 1;
if (i % pri[j] == 0)
{
miu[i * pri[j]] = 0;
break;
}
else
miu[i * pri[j]] = -miu[i];
}
}
for (int i = 1; i <= N - 10; i++)
sum[i] = sum[i - 1] + miu[i];
}
ll Ans(int n, int m)
{
ll ans = 0;
if(n > m)
{
int c = n; n = m; m = c;
}
n /= d, m /= d;
for (int l = 1, r; l <= n; l = r + 1)
{
r = min (n / (n / l), m / (m / l));
ans += (sum[r] - sum[l - 1]) * (n / l) * (m / l);
}
return ans;
}
int main()
{
prework();
for (scanf ("%lld", &t); t--; )
{
scanf("%lld%lld%lld%lld%lld", &n, &m, &n1, &m1, &d);
printf("%lld
", Ans(m, m1) - Ans(m, n1 - 1) - Ans(n - 1, m1) + Ans(n1 - 1, n - 1));
}
return 0;
}