T1、跳房子
\(Sol\):
可以单独考虑一个点进行两种不同的操作后对答案的贡献,随便推一推就能算了。
或者可以把行列的操作分开算,不难发现行的和是个等差数列(列也一样),只需要先进行(xing)行(hang)操作,维护每一列的和的首项和公差即可;
时间复杂度 \(O(m)\)。
代码如下:
//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include <cstdio>
#include <cstring>
#include <algorithm>
int in() {
int x = 0; char c = getchar(); bool f = 0;
while (c < '0' || c > '9')
f |= c == '-', c = getchar();
while (c >= '0' && c <= '9')
x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
return f ? -x : x;
}
template<typename T>inline void chk1(T &_, T __) { _ = _ < __ ? _ : __; }
template<typename T>inline void chk2(T &_, T __) { _ = _ > __ ? _ : __; }
const int N = 1e6 + 5, Q = 1e5 + 5, mod = 1e9 + 7;
int n, m, q;
struct info {
int typ;
int x, k;
} a[Q];
int x[N], y[N];
inline void add(int &_, int __) {
_ += __;
if (_ >= mod) _ -= mod;
if (_ < 0) _ += mod;
}
int main() {
//freopen("in", "r", stdin);
freopen("game.in", "r", stdin);
freopen("game.out", "w", stdout);
n = in(), m = in(), q = in();
for (int i = 1; i <= q; ++i) {
char c = getchar();
while (c != 'R' && c != 'S')
c = getchar();
a[i] = (info){c == 'R', in(), in()};
}
int d = n;
for (int i = 1; i <= n; ++i)
x[i] = 1;
if (n & 1)
y[1] = 1ll * (1ll * (n - 1) / 2 * m + 1) % mod * n % mod;
else
y[1] = 1ll * (1ll * (n - 1) * m + 2) % mod * (n / 2) % mod;
for (int i = 1; i <= q; ++i)
if (a[i].typ) {
add(y[1], 1ll * (1 + 1ll * (a[i].x - 1) * m % mod) * x[a[i].x] % mod * (a[i].k - 1) % mod);
add(d, 1ll * (a[i].k - 1) * x[a[i].x] % mod);
x[a[i].x] = 1ll * x[a[i].x] * a[i].k % mod;
}
for (int i = 2; i <= m; ++i)
y[i] = y[i - 1], add(y[i], d);
for (int i = 1; i <= q; ++i)
if (!a[i].typ)
y[a[i].x] = 1ll * y[a[i].x] * a[i].k % mod;
for (int i = 1; i <= m; ++i)
add(y[0], y[i]);
printf("%d\n", y[0]);
return 0;
}
T2、跳房子
\(Sol\):
考虑优化暴力找循环节,记录每个点会走到第 \(m\) 列的哪一个位置,修改时找到第一列中会改变的区间即可;
时间复杂度 \(O((m + n)\ Q)\)。
代码(咕)
T3、优美序列
先咕着,