poj3685 Matrix

思路：

二分套二分。

矩阵在每一列上是严格递增的，可以利用这一点进行二分。

实现：

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll n, m;
ll cal(ll i, ll j)
{
    return i * i + i * 100000 + j * j - j * 100000 + i * j;
}
bool check(ll x)
{
    ll sum = 0;
    for (int i = 1; i <= n; i++)
    {
        int l = 1, r = n, ans = 0;
        while (l <= r)
        {
            int m = l + r >> 1;
            ll tmp = cal(m, i);
            if (tmp <= x) { ans = m; l = m + 1; }
            else r = m - 1;
        }
        sum += ans;
    }
    return sum >= m;
}
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%lld %lld", &n, &m);
        ll l = INF, r = -INF, ans = -INF;
        for (int i = 1; i <= n; i++)
        {
            l = min(l, cal(1, i));
            r = max(r, cal(n, i));
        }
        while (l <= r)
        {
            ll m = l + r >> 1;
            if (check(m)) { ans = m; r = m - 1; }
            else l = m + 1;
        }
        printf("%lld
", ans);
    }
    return 0;
}