题意：有h×w大的公告板。有n条公告要写入，每条公告高度都是1，宽度是wi，每次从最上最左的空位写，假设有空位输出第几行。假设没有足够空位输出-1。

题解：注意h最大1e9。但事实上是看n的大小。由于假设有n条公告最多占n行，所以线段树最大是min(h,n)。

#include 
         
          #include 
          
           #include 
           
            using namespace std;const int N = 200005;int h, w, n, sum[N << 2];void pushup(int k) {    sum[k] = max(sum[k * 2], sum[k * 2 + 1]);}void build(int k, int left, int right) {    if (left == right) {        sum[k] = w;        return;    }    int mid = (left + right) / 2;    build(k * 2, left, mid);    build(k * 2 + 1, mid + 1, right);    pushup(k);}int query(int k, int left, int right, int x) {    if (left == right) {        sum[k] -= x;        return left;    }    int mid = (left + right) / 2, res = -1;    if (sum[k * 2] >= x)        res = query(k * 2, left, mid, x);    else if (sum[k * 2 + 1] >= x)        res = query(k * 2 + 1, mid + 1, right, x);    pushup(k);    return res;}int main() {    while (scanf("%d%d%d", &h, &w, &n) == 3) {        int temp = h < n ? h : n;        build(1, 1, temp);        int x;        for (int i = 0; i < n; i++) {            scanf("%d", &x);            if (x > w || sum[1] < x)                printf("-1\n");            else                printf("%d\n", query(1, 1, temp, x));        }    }    return 0;}