NOIP模拟70

T3：

　　考虑首先显然可以按行分别考虑，问题为如何快速计算一段区间的贡献

考虑合并区间，然而并没有发现正确的合并策略，于是考虑优化暴力，利用倍增

发现一个显然的贪心策略为由于每个位置比选，于是从区间第一个x开始k长度贪心选择即可

发现枚举x位置的过程可以进行倍增优化，设计dp转移每个位置往后2^j个k区间即可

代码如下：

#include <bits/stdc++.h>
using namespace std;
#define I int
#define C char
#define B bool
#define V void
#define LL long long
#define UI unsigned int
#define UL unsigned long long
#define P pair<I,I>
#define MP make_pair
#define a first
#define b second
#define lowbit(x) (x & -x)
#define debug cout << "It's Ok Here !" << endl;
#define FP(x) freopen (#x,"r",stdin)
#define FC(x) freopen (#x,"w",stdout)
const I N = 1e5 + 3;
I H,W,K,Q,D,L,R,tmp[N],pos[15][N][20];
C s[15][N];
inline I read () {
    I x(0),y(1); C z(getchar());
    while (!isdigit(z)) { if (z == '-') y = -1; z = getchar(); }
    while ( isdigit(z))  x = x * 10 + (z ^ 48), z = getchar();
    return x * y;
}
inline V Max (I &a,I b) { a = a > b ? a : b; }
inline V Min (I &a,I b) { a = a < b ? a : b; }
inline I max (I a,I b) { return a > b ? a : b; }
inline I min (I a,I b) { return a < b ? a : b; }
inline V swap (I &a,I &b) { a ^= b, b ^= a, a ^= b; }
inline I abs (I a) { return a >= 0 ? a : -a; }
inline P operator + (const P &a,const P &b) {
    return MP (a.a + b.a,a.b + b.b);
}
inline P operator - (const P &a,const P &b) {
    return MP (a.a - b.a,a.b - b.b);
}

signed main () {
    FP (blueshit.in), FC (blueshit.out);
    H = read (), W = read (), K = read (), Q = read ();
    for (I i(1);i <= W; ++ i)
        tmp[i] = log2 (i);
    for (I i(1);i <= H; ++ i) {
        scanf ("%s",s[i] + 1);
        for (I j(1);j <= W; ++ j) {
            pos[i][j][0] = s[i][j] == 'X' ? 
                j : pos[i][j - 1][0];
            for (I k(1);k <= tmp[j]; ++ k) 
                pos[i][j][k] = pos[i][max (pos[i][j][k - 1] - K,0)][k - 1];
        }
    }
    while (Q -- ) {
        I ans (0);
        D = read (), L = read (), R = read ();
        for (I i(1);i <= D; ++ i) {
            I it (R), len (R - L + 1);
            for (I j(tmp[len]); ~j ; -- j) 
                if (pos[i][it][j] >= L)
                    ans += 1 << j, it = max (pos[i][it][j] - K,0);
        }
        printf ("%d\n",ans);
    }
}