标签：play cto 根据 odi class 恶心 phi 注释 str
一口气做完六个省的省选（误）
枚举最大的天数，然后代价贪心地O(1)计算。
1 #include <cstdio> 2 #include <algorithm> 3 4 #define R register 5 typedef long long ll; 6 #define maxn 100010 7 #define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0) 8 #define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0) 9 #define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b)) 10 int a[maxn], b[maxn]; 11 int main() 12 { 13 R ll A, B, C, sum = 0, suma = 0; scanf("%lld%lld%lld", &A, &B, &C); 14 R int n, m, maxx = 0; scanf("%d%d", &n, &m); 15 for (R int i = 1; i <= n; ++i) scanf("%d", &a[i]), suma += a[i]; 16 for (R int i = 1; i <= m; ++i) scanf("%d", &b[i]), sum += b[i]; 17 std::sort(a + 1, a + n + 1); 18 std::sort(b + 1, b + m + 1); 19 maxx = b[m]; 20 R int p = m + 1, pp = n + 1; 21 R ll pre = sum, suf = 0, ans = 1e18; 22 for (R int i = maxx; i; --i) 23 { 24 while (p && b[p - 1] >= i) --p, suf += b[p], pre -= b[p]; 25 while (pp && a[pp - 1] >= i) suma -= a[--pp]; 26 R ll need = suf - 1ll * i * (m - p + 1); 27 if (A >= B) 28 { 29 R ll cost = (1ll * i * (pp - 1) - suma) * C + need * B; 30 cost > 0 ? cmin(ans, cost) : 0; 31 } 32 else 33 { 34 R ll v = 1ll * i * (p - 1) - pre; 35 // printf("v = %lld %lld %d\n", v, need, i); 36 // printf("pp %d suma %lld\n", pp, suma); 37 cmin(v, need); 38 R ll cost = (1ll * i * (pp - 1) - suma) * C + v * A; 39 need -= v; 40 cost += need * B; 41 cost > 0 ? cmin(ans, cost) : 0; 42 // printf("%lld\n", ans); 43 } 44 } 45 printf("%lld\n", ans); 46 return 0; 47 }
恶心题。广义欧拉定理+线段树。
不懂什么是广义欧拉定理的同学（比如我）自行Baidu吧。大意和普通的欧拉定理是差不多的。
我们想象一下，指数第一次是模phi(p)，第二次就是模phi(phi(p))，如此迭代下去直到模数变成1，那么就和原先的数无关了，并且因为后面都是模1，到最后的时候每次操作相当于最上面的那个c拿去模1，结果还是c^c^c...的多少次方，所以它的值没有变。
一直迭代phi的这个操作我们联想到了 2749: [HAOI2012]外星人 ，可以证明这个操作次数是log级别的（详见那题题解）。
根据以上预备知识我们可以得到以下算法：
用线段树维护区间没有变成自环的点的位置，然后每次找到这个点暴力重新计算数值。
复杂度是每个点最多算log次*每次要算log次*快速幂的log，所以是3个log的。（很爆炸的复杂度）
1 #include <cstdio> 2 3 #define R register 4 #define maxn 50010 5 #define maxh 25 6 #define maxx 500000 7 typedef long long ll; 8 bool flag, fl[maxh][maxx + 10], vis[maxh][maxx + 10]; 9 int dp[maxh][maxx + 10]; 10 int ph[2333], phcnt, fc, p, c; 11 int counter; 12 inline int qpow(R int base, R int power, R int mh) 13 { 14 flag = 0; 15 R bool memo = power <= maxx && phcnt - mh < maxh; 16 if (memo && vis[phcnt - mh][power]) {flag = fl[phcnt - mh][power]; ++counter; return dp[phcnt - mh][power];} 17 R int mod = ph[mh], tp = power; 18 // fprintf(stderr, "base %d %d\n", mh, power); 19 R int ret = 1; ret >= mod ? flag = 1, ret %= mod : 0; 20 for (R ll t; power; power >>= 1, base = 1ll * base * base % mod) 21 power & 1 ? (t = 1ll * ret * base) >= mod ? flag = 1, ret = t % mod : ret = t : 0; 22 // printf("flag %d\n", flag); 23 memo ? vis[phcnt - mh][tp] = 1, fl[phcnt - mh][tp] = flag, dp[phcnt - mh][tp] = ret : 0; 24 return ret; 25 } 26 inline int phi(R int x) 27 { 28 R int t = x, ret = 1; 29 // printf("%d\n", x); 30 for (R int i = 2; 1ll * i * i <= x; ++i) 31 if (t % i == 0) 32 { 33 t /= i; ret *= i - 1; 34 while (t > 1 && t % i == 0) 35 t /= i, ret *= i; 36 } 37 t != 1 ? ret *= t - 1 : 0; 38 return ret; 39 } 40 ll sum[maxn << 2]; 41 bool tr[maxn << 2]; 42 int cov[maxn], a[maxn], ql, qr, qv; 43 inline void update(R int o) 44 { 45 sum[o] = (sum[o << 1] + sum[o << 1 | 1]) % p; 46 tr[o] = tr[o << 1] && tr[o << 1 | 1]; 47 } 48 void build(R int o, R int l, R int r) 49 { 50 if (l == r) 51 { 52 sum[o] = a[l]; 53 tr[o] = phcnt == 0; 54 return ; 55 } 56 R int mid = l + r >> 1; 57 build(o << 1, l, mid); build(o << 1 | 1, mid + 1, r); 58 update(o); 59 } 60 int query(R int o, R int l, R int r) 61 { 62 if (ql <= l && r <= qr) return sum[o]; 63 R int mid = l + r >> 1, ret = 0; 64 if (ql <= mid) (ret += query(o << 1, l, mid)) %= p; 65 if (mid < qr) (ret += query(o << 1 | 1, mid + 1, r)) %= p; 66 return ret; 67 } 68 void modify(R int o, R int l, R int r) 69 { 70 if (l == r) 71 { 72 if (tr[o]) return ; 73 ++cov[l]; 74 a[l] >= ph[cov[l]] ? flag = 1 : flag = 0; 75 R int ff = a[l] % ph[cov[l]]; 76 // printf("apos %d flag %d\n", a[l], flag); 77 for (R int i = cov[l]; i; --i) ff = qpow(c, ff + (flag ? ph[i] : 0), i - 1); 78 79 // printf("l %d ff %d\n", l, ff); 80 tr[o] = cov[l] == phcnt; 81 sum[o] = ff; 82 return ; 83 } 84 R int mid = l + r >> 1; 85 if (ql <= l && r <= qr) 86 { 87 if (!tr[o << 1]) modify(o << 1, l, mid); 88 if (!tr[o << 1 | 1]) modify(o << 1 | 1, mid + 1, r); 89 } 90 else 91 { 92 if (ql <= mid) modify(o << 1, l, mid); 93 if (mid < qr) modify(o << 1 | 1, mid + 1, r); 94 } 95 update(o); 96 } 97 int main() 98 { 99 R int n, m; scanf("%d%d%d%d", &n, &m, &p, &c); 100 ph[0] = p; for (; ph[phcnt] != 1;) ph[++phcnt] = phi(ph[phcnt - 1]); 101 ph[++phcnt] = 1; 102 // fprintf(stderr, "phcnt %d\n", phcnt); 103 // for (R int i = phcnt - 1; i; --i) fc = qpow(c, fc + ph[i], ph[i - 1]); 104 // printf("%d\n", fc); 105 for (R int i = 1; i <= n; ++i) scanf("%d", &a[i]); 106 build(1, 1, n); 107 for (; m; --m) 108 { 109 R int opt, l, r; scanf("%d%d%d", &opt, &l, &r); 110 if (opt == 0) 111 { 112 ql = l; qr = r; 113 modify(1, 1, n); 114 } 115 else 116 { 117 ql = l; qr = r; 118 printf("%d\n", query(1, 1, n)); 119 } 120 } 121 // fprintf(stderr, "%d\n", counter); 122 return 0; 123 }
循环矩阵快速幂，做法和 4818: [Sdoi2017]序列计数 有点像。
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 5 #define R register 6 typedef long long ll; 7 int k, p; 8 typedef int Vector[60]; 9 Vector base, ans; 10 void mul(R Vector A, R Vector B) 11 { 12 R Vector C; memset(C, 0, k << 2); 13 for (R int i = 0; i < k; ++i) 14 for (R int j = 0; j < k; ++j) 15 C[(i + j) % k] = (C[(i + j) % k] + 1ll * A[i] * B[j]) % p; 16 memcpy(A, C, k << 2); 17 } 18 int main() 19 { 20 R int n, r; scanf("%d%d%d%d", &n, &p, &k, &r); 21 base[0] = 1; ++base[k - 1]; 22 ans[0] = 1; 23 for (R ll power = 1ll * n * k; power; power >>= 1, mul(base, base)) 24 power & 1 ? mul(ans, base), 1 : 0; 25 // for (R int i = 0; i < k; ++i) printf("%d ", ans[i]); 26 printf("%d\n", ans[r]); 27 return 0; 28 }
恶心题*2。树形DP
这题本来写了一个换根的做法结果因为太乱太复杂就推掉重新写了。（论想清楚再写的重要性）
最后参考了 SD_le 做法的题解。这位神犇的状态设计有理有据令人信服，所以我就参考了他的DP状态设计，然后自己推了一遍转移。然而推完以后还是不能1A，对拍以后才发现自己漏掉了好几种情况。。。（我好菜啊.jpg）
1 #include <cstdio> 2 #include <cstring> 3 4 #define R register 5 #define maxn 500010 6 struct Edge { 7 Edge *next; 8 int to; 9 } *last[maxn], e[maxn << 1], *ecnt = e; 10 inline void link(R int a, R int b) 11 { 12 *++ecnt = (Edge) {last[a], b}; last[a] = ecnt; 13 *++ecnt = (Edge) {last[b], a}; last[b] = ecnt; 14 } 15 int f1[maxn], f2[maxn], f3[maxn], f4[maxn], g1[maxn], g2[maxn], ans; 16 #define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0) 17 void dfs(R int x, R int fa) 18 { 19 R int tf1, tf2, tf3, tf4, tg1, tg2, deg = 0, mx = 0; 20 R bool debug = x == 1; 21 for (R Edge *iter = last[x]; iter; iter = iter -> next) 22 if (iter -> to != fa) 23 { 24 dfs(iter -> to, x); ++deg; 25 tf1 = f1[x]; tf2 = f2[x]; tf3 = f3[x]; tf4 = f4[x]; tg1 = g1[x]; tg2 = g2[x]; 26 27 cmax(tg1, f2[iter -> to]); 28 cmax(tg1, g1[iter -> to] - 1); 29 30 cmax(tg2, g1[x] + g1[iter -> to] - 1); 31 cmax(tg2, g1[x] + f2[iter -> to]); 32 cmax(tg2, f4[iter -> to]); 33 cmax(tg2, g2[iter -> to] - 1); 34 35 36 cmax(tf1, f1[iter -> to] - 1); 37 38 cmax(tf2, f1[x] + f1[iter -> to] - 1); 39 40 cmax(tf3, f1[x] + f2[iter -> to] - 1);// && debug ? printf("t1 %d %d\n", iter -> to, tf3) : 0; 41 cmax(tf3, f1[x] + g1[iter -> to] - 1);// && debug ? printf("t2 %d %d\n", iter -> to, tf3) : 0; 42 cmax(tf3, f2[x] + f1[iter -> to] - 1);// && debug ? printf("t3 %d %d\n", iter -> to, tf3) : 0; 43 cmax(tf3, mx + f1[iter -> to] - 2);// && debug ? printf("t4 %d %d %d\n", iter -> to, tf3) : 0; 44 cmax(tf3, f3[iter -> to] - 1);// && debug ? printf("t5 %d %d\n", iter -> to, tf3) : 0; 45 46 cmax(tf4, f1[x] + f3[iter -> to] - 1); 47 cmax(tf4, f3[x] + f1[iter -> to] - 1); 48 cmax(tf4, f2[x] + g1[iter -> to] - 1); 49 cmax(tf4, f2[x] + f2[iter -> to] - 1); 50 51 cmax(mx, f2[iter -> to]); 52 cmax(mx, g1[iter -> to]); 53 f1[x] = tf1; f2[x] = tf2; f3[x] = tf3; f4[x] = tf4; g1[x] = tg1; g2[x] = tg2; 54 // printf("f %d %d %d %d iter -> to %d %d\n", tf1, tf2, tf3, tf4, iter -> to, x); 55 } 56 ++g1[x]; ++g2[x]; 57 f1[x] += deg; f2[x] += deg; f3[x] += deg; f4[x] += deg; 58 /*cmax(g1[x], 1);*/ cmax(g2[x], g1[x]); 59 cmax(f2[x], f1[x]); cmax(f3[x], f2[x]); cmax(f4[x], f3[x]); 60 cmax(ans, g2[x]); cmax(ans, f4[x]); 61 // printf("ans %d x %d\n", ans, x); 62 // x == 2 ? printf("f %d %d %d %d\ng %d %d\n", f1[x], f2[x], f3[x], f4[x], g1[x], g2[x]) : 0; 63 } 64 int main() 65 { 66 R int T, type; scanf("%d%d", &T, &type); 67 for (; T; --T) 68 { 69 R int n; scanf("%d", &n); ans = 0; 70 for (R int i = 1; i <= type; ++i) scanf("%*d%*d"); 71 for (R int i = 1; i < n; ++i) 72 { 73 R int a, b; scanf("%d%d", &a, &b); link(a, b); 74 } 75 if (n == 1) {puts("0"); continue;} 76 dfs(1, 0); 77 printf("%d\n", ans); 78 79 memset(last, 0, (n + 1) << 2); ecnt = e; 80 memset(f1, 0, (n + 1) << 2); 81 memset(f2, 0, (n + 1) << 2); 82 memset(f3, 0, (n + 1) << 2); 83 memset(f4, 0, (n + 1) << 2); 84 memset(g1, 0, (n + 1) << 2); 85 memset(g2, 0, (n + 1) << 2); 86 } 87 return 0; 88 } 89 /* 90 1 0 91 15 92 1 2 93 2 3 94 1 4 95 2 5 96 1 6 97 3 7 98 4 8 99 3 9 100 4 10 101 6 11 102 5 12 103 4 13 104 6 14 105 5 15 106 9 107 108 1 0 109 10 110 1 2 111 2 3 112 1 4 113 3 5 114 3 6 115 2 7 116 1 8 117 7 9 118 9 10 119 6 120 121 1 0 122 10 123 1 2 124 1 3 125 2 4 126 1 5 127 3 6 128 3 7 129 5 8 130 5 9 131 4 10 132 */
高斯消元有95分。
先把倍数关系通过一次高斯消元转化为单个灯是否要改变。
计c[i]表示还剩下i个1时的期望步数。根据题意，对于i<=k,有c[i] = i。然后这个转移有环，所以得高斯消元。
推出来每一行只有3个变量，所以消元是线性的。因为n和模数很接近，而消元的时候又有+1，然后系数很容易就+到0了，然后就boom（爆炸熊.jpg）。
既然出题人卡了高斯消元那么我们就来考虑推推式子。（因为没有用markdown所以式子略丑请原谅。。。）
考虑我们之前推出来的递推式：c[i] = (i/n) * c[i-1] + ((n - i) / n) * c[i + 1] + 1
然后对于i=n的时候，有c[n] = c[n - 1] + 1 (1)
对于i=n-1时，有c[n-1] = 1 + ((n - 1) / n) * c[n - 2] + (1 / n) * c[n] (2)
把(1)式代入(2)式：c[n - 1] = 1 + ((n - 1) / n) * c[n - 2] + 1 / n + 1 / n * c[n - 1]
移项：((n - 1) / n) * c[n - 1] = ((n - 1) / n) * c[n - 2] + (n + 1) / n
除过去：c[n - 1] = c[n - 2] + (n + 1) / (n - 1)
只剩两项啦~于是我们猜想c[i]这个序列是一个一阶递推式，并且上一项的系数是1（这个可以用数学归纳法证）。
于是我们就只要知道递推的常数项即可。我们来强行分析一波：
设第i项的系数为x[i]，而且我们已知了x[n] = 1。递推式可以写成c[i] = c[i - 1] + x[i]
我们从一开始的式子入手：c[i] = (i / n) * c[i - 1] + ((n - i) / n) * c[i + 1] + 1
代入：c[i] = (i / n) * c[i - 1] + ((n - i) / n) * (c[i] + x[i + 1]) + 1
两边乘个n然后再移项：i * c[i] = i * c[i - 1] + (n - i) * x[i + 1] + n
除个i：c[i] = c[i - 1] + ((n - i) * x[i + 1] + n) / i
这样的话连上面系数等于1的顺便也给证了。。。
所以求得x[i] = ((n - i) * x[i + 1] + n) / i。
然后递推一下就能求出c每一项的值来啦。
附代码：（95分的高斯消元在注释里面）
1 #include <cstdio> 2 #define R register 3 #define maxn 100010 4 const int mod = 1e5 + 3; 5 int a[maxn], b[maxn], c[maxn][3], f[maxn], x[maxn], inv[maxn]; 6 inline int qpow(R int base, R int power) 7 { 8 R int ret = 1; 9 for (; power; power >>= 1, base = 1ll * base * base % mod) 10 power & 1 ? ret = 1ll * ret * base % mod : 0; 11 return ret; 12 } 13 int main() 14 { 15 R int n, k, fact = 1; scanf("%d%d", &n, &k); 16 for (R int i = 1; i <= n; ++i) scanf("%d", &a[i]); 17 inv[1] = 1; 18 for (R int i = 2; i <= n; ++i) 19 { 20 fact = 1ll * fact * i % mod, inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod; 21 // printf("%d %d\n", inv[i], qpow(i, mod - 2)); 22 } 23 R int tmp = k, kk = 0; 24 for (R int i = n; i; --i) 25 { 26 b[i] = a[i]; 27 for (R int j = i << 1; j <= n; j += i) b[i] ^= b[j]; 28 kk += b[i]; 29 } 30 if (kk <= k) return !printf("%d\n", 1ll * kk * fact % mod); 31 /* c[k][0] = 1; c[k][1] = 0; c[k][2] = k; 32 for (R int i = k + 1; i <= n; ++i) 33 { 34 R int tmp = 1ll * qpow(c[i - 1][0], mod - 2) * i % mod * qpow(n, mod - 2) % mod; 35 c[i][0] = (1 + 1ll * c[i - 1][1] * tmp) % mod; 36 c[i][1] = 1ll * (mod + i - n) * qpow(n, mod - 2) % mod; 37 c[i][2] = (1 + 1ll * c[i - 1][2] * tmp) % mod; 38 } 39 for (R int i = n; i > k; --i) 40 { 41 f[i] = 1ll * (c[i][2] - 1ll * f[i + 1] * c[i][1] % mod + mod) * qpow(c[i][0], mod - 2) % mod; 42 } 43 printf("%d\n", 1ll * f[kk] * fact % mod);*/ 44 x[n] = 1; 45 R int ans = k; 46 for (R int i = n - 1; i > k; --i) 47 { 48 x[i] = (1ll * x[i + 1] * (n - i) % mod * inv[n] % mod + 1) * n % mod * inv[i] % mod; 49 } 50 for (R int i = k + 1; i <= kk; ++i) (ans += x[i]) %= mod; 51 printf("%d\n", 1ll * ans * fact % mod); 52 return 0; 53 }
最大权闭合子图。最小割。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 5 #define R register 6 #define maxn 100010 7 #define inf 0x7fffffff 8 #define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b)) 9 struct Edge { 10 Edge *next, *rev; 11 int to, cap; 12 } *last[maxn], *cur[maxn], e[maxn << 2], *ecnt = e; 13 inline void link(R int a, R int b, R int w) 14 { 15 // printf("%d %d %d\n", a, b, w); 16 *++ecnt = (Edge) {last[a], ecnt + 1, b, w}; last[a] = ecnt; 17 *++ecnt = (Edge) {last[b], ecnt - 1, a, 0}; last[b] = ecnt; 18 } 19 int a[110], d[110][110], id1[1010], id2[1010], id[110][110]; 20 int ans, s, t, q[maxn], dep[maxn]; 21 inline bool bfs() 22 { 23 memset(dep, -1, (t + 1) << 2); 24 dep[q[1] = t] = 0; R int head = 0, tail = 1; 25 while (head < tail) 26 { 27 R int now = q[++head]; 28 for (R Edge *iter = last[now]; iter; iter = iter -> next) 29 if (dep[iter -> to] == -1 && iter -> rev -> cap) 30 dep[q[++tail] = iter -> to] = dep[now] + 1; 31 } 32 return dep[s] != -1; 33 } 34 int dfs(R int x, R int f) 35 { 36 if (x == t) return f; 37 R int used = 0; 38 for (R Edge* &iter = cur[x]; iter; iter = iter -> next) 39 if (iter -> cap && dep[iter -> to] + 1 == dep[x]) 40 { 41 R int v = dfs(iter -> to, dmin(f - used, iter -> cap)); 42 iter -> cap -= v; 43 iter -> rev -> cap += v; 44 used += v; 45 if (used == f) return f; 46 } 47 return used; 48 } 49 inline void dinic() 50 { 51 while (bfs()) 52 { 53 memcpy(cur, last, sizeof cur); 54 ans += dfs(s, inf); 55 } 56 } 57 int main() 58 { 59 R int n, m, sum = 0; scanf("%d%d", &n, &m); 60 for (R int i = 1; i <= n; ++i) scanf("%d", &a[i]); 61 for (R int i = 1; i <= n; ++i) for (R int j = i; j <= n; ++j) scanf("%d", &d[i][j]); 62 R int tot = 0; 63 for (R int i = 1; i <= n; ++i) 64 { 65 for (R int j = i; j <= n; ++j) 66 id[i][j] = ++tot; 67 link(id[i][i], id1[a[i]] ? id1[a[i]] : id1[a[i]] = ++tot, a[i]); 68 m ? link(id[i][i], id2[a[i]] ? id2[a[i]] : id2[a[i]] = ++tot, inf), 1 : 0; 69 } 70 t = ++tot; 71 for (R int i = 1; i <= n; ++i) 72 { 73 for (R int j = i; j <= n; ++j) 74 { 75 d[i][j] > 0 ? link(s, id[i][j], d[i][j]), sum += d[i][j] : (link(id[i][j], t, -d[i][j]), 0); 76 i != j ? link(id[i][j], id[i + 1][j], inf), link(id[i][j], id[i][j - 1], inf), 1 : 0; 77 } 78 } 79 for (R int i = 0; i <= 1000; ++i) 80 { 81 id1[i] ? link(id1[i], t, inf), 1 : 0; 82 id2[i] ? link(id2[i], t, i * i), 1 : 0; 83 } 84 dinic(); 85 // printf("ans = %d %d\n", ans, t); 86 printf("%d\n", sum - ans); 87 return 0; 88 }
标签：play cto 根据 odi class 恶心 phi 注释 str
原文地址：http://www.cnblogs.com/cocottt/p/6819036.html
