Day3-P - Matrix POJ3685

Given a N × N matrix A, whose element in the i-th row and j-th column A_ij is an number that equals i² + 100000 × i + j² - 100000 × j + i × j, you are to find the M-th smallest element in the matrix.

Input

The first line of input is the number of test case.
For each test case there is only one line contains two integers, N(1 ≤ N ≤ 50,000) and M(1 ≤ M ≤ N × N). There is a blank line before each test case.

Output

For each test case output the answer on a single line.

Sample Input

12

1 1

2 1

2 2

2 3

2 4

3 1

3 2

3 8

3 9

5 1

5 25

5 10

Sample Output

3
-99993
3
12
100007
-199987
-99993
100019
200013
-399969
400031
-99939

思路：观察列递增，行递减，枚举列进行二分算出第M大，然后二分枚举X是第几大，二分套二分，代码如下：

typedef long long LL;

const LL INF = 0x3f3f3f3f3f3f3;

LL N, M;

LL calculate(LL i, LL j) {
    return i * i + 100000 * i + j * j - 100000 * j + i * j;
}

LL check(LL num) {
    LL sum = 0, l, r, mid;
    for (int i = 1; i <= N; ++i) {
        l = 1, r = N;
        while(l <= r) {
            mid = (r + l) >> 1;
            if(calculate(mid,i) > num) {
                r = mid - 1;
            }
            else
                l = mid + 1;
        }
        sum += r;
    }
    return sum;
}

int main() {
    LL T;
    scanf("%I64d", &T);
    while(T--) {
        scanf("%I64d%I64d", &N, &M);
        LL l = -INF, r = INF, mid;
        while(l <= r) {
            mid = (r + l) >> 1;
            if(check(mid) < M) {
                l = mid + 1;
            }
            else
                r = mid - 1;
        }
        printf("%I64d\n", l);
    }
    return 0;
}

　小结：如何寻找二分的答案，如果mid是所求答案，那么check(mid)为true，更新r的值为mid-1，之后的check都是false，从而更新l，最后l变为mid，输出mid。

　比如>=的情况就是true的情况，所以一般答案都是非>=的那一侧。

补：
在二分时，注意满足的条件是满足题意还是不满足题意，New如下：(一般来说ans都在等于的那一侧，但是哪一侧是l哪一侧是r需要判断

typedef long long LL;

LL N, M, mid, ans;

LL calculate(LL i, LL j) {
    return i * i + 100000 * i + j * j - 100000 * j + i * j;
}

LL check(LL x) {
    LL l, r, mid, sum = 0, ans;
    for(int i = 1; i <= N; ++i) {
        l = 1, r = N, ans = 0;
        while(l <= r) {
            mid = (r + l) >> 1;
            if(calculate(mid, i) <= x) {
                ans = mid;
                l = mid + 1;
            } else {
                r = mid - 1;
            }
                
        }
        sum += ans;
    }
    return sum;
}

const LL INF = 0x3f3f3f3f3f3f3;
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%I64d%I64d", &N, &M);
        //LL l = calculate(1, N), r = calculate(N, 1);
        LL l = -INF, r = INF;
        while(l <= r) {
            mid = (r + l) >> 1;
            if(check(mid) < M) {
                l = mid + 1;
            } else {
                ans = mid;
                r = mid - 1;
            }
        }
        printf("%I64d\n", ans);
    }
    return 0;
}