T1 药品实验
内网#4803
由概率定义，有$$a + b + c = 0$$
变形得到$$1 - b = a + c$$
根据题意有$$p_i = a p {i - 1} + b p_i + c p{i + 1}$$
[ herefore (1 - b) p_i = a p_{i - 1} + c p_{i + 1}
]
[ herefore (a + c) p_i = a p_{i - 1} + c p_{i + 1}
]
[a(p_i - p_{i - 1}) = c(p_{i + 1} - p_i)
]
仔细观察一下发现这对于任何(i)都适用，也就是说(p)序列的差分序列有一些优秀的性质
设(d)为(p)的差分序列，(k)为(d)的公比即(frac{a}{c})，那么有$$a* d_i = c * d_{i + 1}$$
即$$frac{d_{i + 1}}{d_i} = frac{a}{c}$$
(d)是一个等比数列，求(p_n)即求(sumlimits_{i = 1}^n d_i)，由等比数列求和公式得到
[p_{2n} = d_1 * frac{(k^n - 1)(k^n + 1)}{k - 1} = 1
]
[p_n = d_1 * frac{1}{k^n + 1} = p_{2n} * frac{1}{k^n + 1} = frac{1}{k^n + 1}
]
(O(log(n)))求快速幂即可
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int mod = (int)1e9 + 7;
ll n, alpha, beta, a, c, ans;
ll qpow(ll a, ll b) {ll ans = 1; for (; b; b >>= 1) {if (b&1) ans = (a * ans) % mod; a = (a * a) % mod;} return ans % mod;}
int main() {
cin >> n >> alpha >> beta;
a = ((1 - alpha) * beta) % mod, c = alpha * (1 - beta) % mod;
ans = qpow(c, mod - 2) * a % mod; ans = qpow(ans, n) % mod + 1; ans = qpow(ans, mod - 2) % mod;
printf("%lld", ans);
return 0;
}
T2 小猫钓鱼
内网#4804
被自己sb到了，(s)没跟着离散化 (100pts -> 10pts)
#include<bits/stdc++.h>
#define N (100000 + 10)
using namespace std;
inline int read(){
int cnt = 0, f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();}
return cnt * f;
}
int T, n, m, s, l, a[N], b[N], cnt, tot, inq[N], kil[N], r;
deque <int> q[205], q2, t;
void pre_work() {
sort (b + 1, b + cnt + 1);
int q = unique(b + 1, b + cnt + 1) - b - 1;
for (register int i = 1; i <= cnt; ++i) a[i] = lower_bound(b + 1, b + q + 1, a[i]) - b;
s = lower_bound(b + 1, b + q + 1, s) - b;
}
int main() {
// freopen("fishing.in", "r", stdin);
// freopen("fishing.out", "w", stdout);
while (1) {
n = read(), m = read(), l = read(), s = read(), T = read();
tot = n; cnt = 0; r = 0;
if (n == -1) break;
memset(inq, 0, sizeof inq);
for (register int i = 1; i <= n; ++i) while (q[i].size()) q[i].pop_back();
while (q2.size()) q2.pop_back();
while (t.size()) t.pop_back();
for (register int i = 1; i <= n; ++i)
for (register int j = 1; j <= l; ++j) a[++cnt] = b[cnt] = read();
a[++cnt] = b[cnt] = s;
pre_work();
for (register int i = 1; i <= n; ++i)
for (register int j = 1; j <= l; ++j) q[i].push_back(a[(i - 1) * l + j]);
while (T-- && tot > 1) { //every round
++r;
for (register int i = 1; i <= n; ++i) { // every person
if (q[i].empty()) continue;
int cur = q[i].front(); q[i].pop_front();
q2.push_back(cur); ++inq[cur];
if (cur == s && q2.size() > 1) { // if cur is J
while (!q2.empty()) t.push_back(q2.back()), --inq[q2.back()], q2.pop_back();
while (!t.empty()) q[i].push_back(t.back()), t.pop_back();
}
if (inq[cur] > 1 && cur != s && q2.size() > 1) {
while (inq[cur] && q2.size()) //pop from back
t.push_back(q2.back()), --inq[q2.back()], q2.pop_back();
while (!t.empty()) q[i].push_back(t.back()), t.pop_back();
}
if (q[i].empty()) --tot, kil[i] = r;
}
}
for (register int i = 1; i <= n; ++i) printf("%d ", q[i].size() ? q[i].size() : -kil[i]);
puts("");
for (register int i = 1; i <= n; ++i) {
while (q[i].size()) {printf("%d ", b[q[i].front()]); q[i].pop_front();}
puts("");
}
}
return 0;
}
T3
太毒瘤了所以咕了