codeforces gym102040 前四题签到题解
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J
Description
输入一个数，输出其1.15倍，保留两位小数
Solution
read&write
E
Description
给出两时间，计算其差值。
Solution
模拟即可
C
Description
给出一个$nleq100000$,求$n!的约数的约数个数$
Solution
数论不会gcd，通过样例模拟了两遍，发现$4!=2^3*3$,素因子分开解决，$2^3时其因子有可能有2^0,2^1,2^2,2^3，而这些数的因子有4*2^0,3*2^1,2*2^2,1*2^3$
嗯?等差数列累加和！！$sum=1+2+3+4=(3+1)(3+2)/2$，那后面素因子3怎么搞，互不相干，分步计算法则，直接乘喽。
1 #include <algorithm> 2 #include <cctype> 3 #include <cmath> 4 #include <cstdio> 5 #include <cstdlib> 6 #include <cstring> 7 #include <iostream> 8 #include <map> 9 #include <queue> 10 #include <set> 11 #include <stack> 12 #if __cplusplus >= 201103L 13 #include <unordered_map> 14 #include <unordered_set> 15 #endif 16 #include <vector> 17 #define lson rt << 1, l, mid 18 #define rson rt << 1 | 1, mid + 1, r 19 #define LONG_LONG_MAX 9223372036854775807LL 20 #define ll LL 21 using namespace std; 22 typedef long long ll; 23 typedef long double ld; 24 typedef unsigned long long ull; 25 typedef pair<int, int> P; 26 int n, m, k; 27 const int maxn = 1e6 + 10; 28 const ll mod = 10000007; 29 const ll inv2 = 7486194; 30 template <class T> 31 inline T read() 32 { 33 int f = 1; 34 T ret = 0; 35 char ch = getchar(); 36 while (!isdigit(ch)) 37 { 38 if (ch == ‘-‘) 39 f = -1; 40 ch = getchar(); 41 } 42 while (isdigit(ch)) 43 { 44 ret = (ret << 1) + (ret << 3) + ch - ‘0‘; 45 ch = getchar(); 46 } 47 ret *= f; 48 return ret; 49 } 50 template <class T> 51 inline void write(T n) 52 { 53 if (n < 0) 54 { 55 putchar(‘-‘); 56 n = -n; 57 } 58 if (n >= 10) 59 { 60 write(n / 10); 61 } 62 putchar(n % 10 + ‘0‘); 63 } 64 template <class T> 65 inline void writeln(const T &n) 66 { 67 write(n); 68 puts(""); 69 } 70 int vis[maxn], prime[maxn], pcnt, num[maxn]; 71 ll qpow(ll a, ll n) 72 { 73 ll res = 1; 74 while (n) 75 { 76 if (n & 1) 77 res = res * a % mod; 78 a = a * a % mod; 79 n >>= 1; 80 } 81 return res; 82 } 83 ll f(ll n, int p)//阶乘因子 84 { 85 if (n == 0) 86 return 0; 87 return f(n / p, p) + n / p; 88 } 89 void init() 90 { 91 for (int i = 2; i < maxn; i++) 92 { 93 if (!vis[i]) 94 prime[pcnt++] = i; 95 for (int j = 0; j < pcnt && i * prime[j] <= maxn; j++) 96 { 97 vis[i * prime[j]] = 1; 98 if (i % prime[j] == 0) 99 break; 100 } 101 } 102 } 103 104 int main(int argc, char const *argv[]) 105 { 106 #ifndef ONLINE_JUDGE 107 // freopen("in.txt", "r", stdin); 108 // freopen("out.txt", "w", stdout); 109 #endif 110 init(); 111 while (~scanf("%d", &n) && n) 112 { 113 memset(num, 0, sizeof(int) * (n + 1)); 114 for (int i = 2; i <= n; i++) 115 if (!vis[i]) 116 num[i] = f(n, i); 117 ll res = 1; 118 for (int i = 2; i <= n; i++) 119 if (num[i]) 120 { 121 if (num[i] & 1) 122 res = res * ((num[i] + 1) >> 1) * (num[i] + 2) % mod; 123 else 124 res = res * ((num[i] + 2) >> 1) * (num[i] + 1) % mod; 125 } 126 writeln(res); 127 } 128 return 0; 129 }
F
Description
给出一个n个点的树，$nleq10000$,m个查询，每次查询k条路径，问k条路径里的公共点个数
Solution
树上路径，先整个树剖。emmm，公共点怎么求？每个点开一个set，最后set求交集，set_intersection,
写好交，wa2。你说超时我都能理解，wa算什么本事。debug了一个小时，还是一直wa，顶不住了，换思路。
我已经把路径存到了线段树里，那么每一条查询边进来的时候路径上的点权值+1，最后查询线段树里权值为k的点不就完了吗(其实想了很久)
每次查询完后清空当前查询加的权值，防止后续影响。
见代码
1 /* 2 gym102040F 树剖+区间覆盖 3 */ 4 #include <algorithm> 5 #include <cctype> 6 #include <cmath> 7 #include <cstdio> 8 #include <cstdlib> 9 #include <cstring> 10 #include <iostream> 11 #include <map> 12 #include <queue> 13 #include <set> 14 #include <stack> 15 #if __cplusplus >= 201103L 16 #include <unordered_map> 17 #include <unordered_set> 18 #endif 19 #include <vector> 20 #define lson rt << 1 21 #define rson rt << 1 | 1 22 #define LONG_LONG_MAX 9223372036854775807LL 23 #define ll LL 24 using namespace std; 25 typedef long long ll; 26 typedef long double ld; 27 typedef unsigned long long ull; 28 typedef pair<int, int> P; 29 int n, m, k, mod; 30 const int maxn = 1e5 + 10; 31 template <class T> 32 inline T read() 33 { 34 int f = 1; 35 T ret = 0; 36 char ch = getchar(); 37 while (!isdigit(ch)) 38 { 39 if (ch == ‘-‘) 40 f = -1; 41 ch = getchar(); 42 } 43 while (isdigit(ch)) 44 { 45 ret = (ret << 1) + (ret << 3) + ch - ‘0‘; 46 ch = getchar(); 47 } 48 ret *= f; 49 return ret; 50 } 51 template <class T> 52 inline void write(T n) 53 { 54 if (n < 0) 55 { 56 putchar(‘-‘); 57 n = -n; 58 } 59 if (n >= 10) 60 { 61 write(n / 10); 62 } 63 putchar(n % 10 + ‘0‘); 64 } 65 template <class T> 66 inline void writeln(const T &n) 67 { 68 write(n); 69 puts(""); 70 } 71 int dfn[maxn], head[maxn], siz[maxn], dep[maxn]; 72 int fa[maxn], top[maxn], a[maxn], son[maxn], w[maxn], cnt, tot, root; 73 struct node 74 { 75 int to, nxt; 76 } edg[maxn << 1]; 77 struct nodeT 78 { 79 int l, r; 80 int lazy, minx, maxx; 81 } tr[maxn << 2]; 82 inline void pushup(int rt) 83 { 84 tr[rt].maxx = max(tr[lson].maxx, tr[rson].maxx); 85 tr[rt].minx = min(tr[lson].minx, tr[rson].minx); 86 } 87 inline void pushdown(int rt) 88 { 89 ll len = tr[rt].r - tr[rt].l + 1; 90 if (tr[rt].lazy) 91 { 92 tr[lson].minx += tr[rt].lazy; //注意此题是区间加，固懒标记也应该是累加而不是覆盖 93 tr[rson].maxx += tr[rt].lazy; 94 tr[lson].maxx += tr[rt].lazy; 95 tr[rson].minx += tr[rt].lazy; 96 tr[lson].lazy += tr[rt].lazy; 97 tr[rson].lazy += tr[rt].lazy; 98 tr[rt].lazy = 0; 99 } 100 } 101 void build(int rt, int l, int r) 102 { 103 tr[rt].l = l; 104 tr[rt].r = r; 105 tr[rt].lazy = 0; 106 tr[rt].maxx = tr[rt].minx = 0; 107 if (l == r) 108 return; 109 int mid = l + r >> 1; 110 build(lson, l, mid); 111 build(rson, mid + 1, r); 112 } 113 void modify(int rt, int L, int R, int v) 114 { 115 int l = tr[rt].l; 116 int r = tr[rt].r; 117 if (l >= L && r <= R) 118 { 119 tr[rt].lazy += v; 120 tr[rt].maxx += v; 121 tr[rt].minx += v; 122 return; 123 } 124 pushdown(rt); //当前区间没有在之前返回代表当前区间并非包含于待查询区间，在向左右区间查询时需要先将懒标记下放 125 int mid = l + r >> 1; 126 if (L <= mid) 127 modify(lson, L, R, v); 128 if (R > mid) 129 modify(rson, L, R, v); 130 pushup(rt); //更新父区间 131 } 132 int query(int rt, int L, int R, int k) 133 { 134 int l = tr[rt].l; 135 int r = tr[rt].r; 136 if (l >= L && r <= R && tr[rt].minx == tr[rt].maxx && tr[rt].minx == k) 137 return r - l + 1; 138 if (l > R || r < L || tr[rt].maxx < k) 139 return 0; 140 pushdown(rt); //和update同理 141 int mid = l + r >> 1; 142 int ans = 0; 143 if (L <= mid) 144 ans += query(lson, L, R, k); 145 if (R > mid) 146 ans += query(rson, L, R, k); 147 return ans; 148 } 149 void add(int x, int y) 150 { 151 edg[tot].nxt = head[x]; 152 edg[tot].to = y; 153 head[x] = tot++; 154 } 155 void dfs1(int u, int p) 156 { 157 fa[u] = p; 158 dep[u] = dep[p] + 1; 159 siz[u] = 1; 160 int maxsz = -1; 161 for (int i = head[u]; ~i; i = edg[i].nxt) 162 { 163 int v = edg[i].to; 164 if (v == p) 165 continue; 166 dfs1(v, u); 167 siz[u] += siz[v]; 168 if (siz[v] > maxsz) 169 { 170 maxsz = siz[v]; 171 son[u] = v; 172 } 173 } 174 } 175 void dfs2(int u, int t) 176 { 177 dfn[u] = ++cnt; 178 top[u] = t; 179 w[cnt] = 0; 180 if (!son[u]) 181 return; 182 dfs2(son[u], t); 183 for (int i = head[u]; ~i; i = edg[i].nxt) 184 { 185 int v = edg[i].to; 186 if (v == fa[u] || v == son[u]) 187 continue; 188 dfs2(v, v); 189 } 190 } 191 inline void mchain(int x, int y, int z) 192 { //修改一条链 193 while (top[x] != top[y]) 194 { 195 if (dep[top[x]] < dep[top[y]]) 196 swap(x, y); 197 modify(1, dfn[top[x]], dfn[x], z); 198 x = fa[top[x]]; 199 } 200 if (dep[x] > dep[y]) 201 swap(x, y); 202 modify(1, dfn[x], dfn[y], z); 203 } 204 int main(int argc, char const *argv[]) 205 { 206 #ifndef ONLINE_JUDGE 207 freopen("in.txt", "r", stdin); 208 #endif 209 int t = read<int>(); 210 for (int cas = 1; cas <= t; cas++) 211 { 212 printf("Case %d: ", cas); 213 n = read<int>(); 214 memset(head, -1, sizeof(int) * (n + 1)); 215 memset(dfn, 0, sizeof(int) * (n + 1)); 216 memset(son, 0, sizeof(int) * (n + 1)); 217 memset(siz, 0, sizeof(int) * (n + 1)); 218 memset(fa, 0, sizeof(int) * (n + 1)); 219 memset(top, 0, sizeof(int) * (n + 1)); 220 memset(dep, 0, sizeof(int) * (n + 1)); 221 memset(w, 0, sizeof(int) * (n + 1)); 222 tot = cnt = 0; 223 for (int i = 1; i < n; i++) 224 { 225 int x = read<int>(), y = read<int>(); 226 add(x, y); 227 add(y, x); 228 } 229 dfs1(1, 0); 230 dfs2(1, 1); 231 build(1, 1, n); 232 m = read<int>(); 233 while (m--) 234 { 235 int k = read<int>(); 236 vector<P> cood; 237 cood.clear(); 238 for (int j = 0; j < k; j++) 239 cood.emplace_back(read<int>(), read<int>()); 240 for (int j = 0; j < k; j++) 241 mchain(cood[j].first, cood[j].second, 1); 242 writeln(query(1, 1, n, k)); 243 for (int j = 0; j < k; j++) 244 mchain(cood[j].first, cood[j].second, -1); 245 } 246 } 247 return 0; 248 }
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