半面数学 半面卡常
1001 Add More Zero
水题不写。
1002 Balala Power!
贪心，注意0不能首位。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 typedef long long LL; 7 const LL mod = 1e9 + 7; 8 const int maxn = 2e5 + 10; 9 int d[26][maxn], sz[26], fir[26], id[26], vis[26]; 10 char s[maxn]; 11 12 bool larger(int i, int j) 13 { 14 if(sz[i] > sz[j]) return true; 15 if(sz[i] < sz[j]) return false; 16 for(int k = sz[i]; k >= 0; k--) 17 if(d[i][k] != d[j][k]) return d[i][k] > d[j][k]; 18 return false; 19 } 20 21 int main(void) 22 { 23 int n, kase = 0; 24 while(~scanf("%d", &n)) 25 { 26 memset(d, 0, sizeof(d)); 27 memset(sz, 0, sizeof(sz)); 28 memset(fir, 0, sizeof(fir)); 29 memset(vis, 0, sizeof(vis)); 30 memset(id, -1, sizeof(id)); 31 for(int i = 1; i <= n; i++) 32 { 33 scanf("%s", s + 1); 34 int l = strlen(s + 1); 35 if(l != 1) fir[s[1]-'a'] = 1; 36 for(int j = l; j; j--) 37 { 38 d[s[j]-'a'][l-j]++; 39 sz[s[j]-'a'] = max(sz[s[j]-'a'], l-j); 40 vis[s[j]-'a'] = 1; 41 } 42 } 43 for(int c = 0; c < 26; c++) 44 for(int i = 0; i <= sz[c]; i++) 45 if(d[c][i] >= 26) d[c][i+1] += d[c][i] / 26, d[c][i] %= 26, sz[c] = max(sz[c], i + 1); 46 int cnt = 0; 47 for(int i = 0; i < 26; i++) cnt += vis[i]; 48 if(cnt == 26) 49 { 50 int p = -1; 51 for(int i = 0; i < 26; i++) 52 { 53 if(fir[i]) continue; 54 if(p == -1 || larger(p, i)) p = i; 55 } 56 id[p] = 0; 57 for(int i = 1; i <= 25; i++) 58 { 59 p = -1; 60 for(int j = 0; j < 26; j++) 61 { 62 if(id[j] != -1) continue; 63 if(p == -1 || larger(p, j)) p = j; 64 } 65 id[p] = i; 66 } 67 } 68 else 69 { 70 for(int i = 25; i >= 26 - cnt; i--) 71 { 72 int p = -1; 73 for(int j = 0; j < 26; j++) 74 { 75 if(id[j] != -1) continue; 76 if(p == -1 || larger(j, p)) p = j; 77 } 78 id[p] = i; 79 } 80 } 81 LL ans = 0; 82 for(int i = 0; i < 26; i++) 83 { 84 if(!vis[i]) continue; 85 LL base = 1; 86 for(int j = 0; j <= sz[i]; j++) ans = (ans + id[i] * base % mod * d[i][j]) % mod, base = base * 26 % mod; 87 } 88 printf("Case #%d: %lld ", ++kase, ans); 89 } 90 return 0; 91 }
1003 Colorful Tree
其实和虚树没什么关系 考虑统计不经过颜色c的路径 端点一定在一些联通块里
对于一个颜色c的节点 遍历子树时把遇到的含有颜色c的子树扣掉就是这个联通块的大小
但是还需要注意的是还会有一个包含root的联通块要单独考虑
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <unordered_map> 6 #include <stack> 7 using namespace std; 8 typedef long long LL; 9 const int maxn = 2e5 + 10; 10 int vis[maxn], c[maxn]; 11 unordered_map< int, stack<int> > st; 12 13 int cnt, h[maxn]; 14 struct edge 15 { 16 int to, pre; 17 } e[maxn<<1]; 18 void add(int from, int to) 19 { 20 cnt++; 21 e[cnt].pre = h[from]; 22 e[cnt].to = to; 23 h[from] = cnt; 24 } 25 26 LL ans; 27 int sz[maxn], num[maxn], rt[maxn]; 28 void dfs(int x, int f) 29 { 30 st[c[x]].push(x); 31 sz[x] = 1, num[x] = 0; 32 for(int i = h[x]; i; i = e[i].pre) 33 { 34 int to = e[i].to; 35 if(to == f) continue; 36 dfs(to, x); 37 sz[x] += sz[to]; 38 ans -= (LL) (sz[to] - num[x]) * (sz[to] - num[x] - 1) / 2; 39 num[x] = 0; 40 } 41 st[c[x]].pop(); 42 if(st[c[x]].empty()) rt[c[x]] += sz[x]; 43 else num[st[c[x]].top()] += sz[x]; 44 } 45 46 int main(void) 47 { 48 int n, kase = 0; 49 while(~scanf("%d", &n)) 50 { 51 st.clear(); 52 cnt = 0; 53 for(int i = 1; i <= n; i++) h[i] = vis[i] = rt[i] = 0; 54 for(int i = 1; i <= n; i++) scanf("%d", c + i), vis[c[i]] = 1; 55 for(int i = 1; i < n; i++) 56 { 57 int u, v; 58 scanf("%d %d", &u, &v); 59 add(u, v), add(v, u); 60 } 61 ans = 0; 62 dfs(1, 0); 63 for(int i = 1; i <= n; i++) if(vis[i]) ans += (LL) n * (n - 1) / 2 - (LL) (sz[1] - rt[i]) * (sz[1] - rt[i] - 1) / 2; 64 printf("Case #%d: %lld ", ++kase, ans); 65 } 66 return 0; 67 }
场上卡了很久的点分治，大概是每次直接统计经过当前根rt的边的贡献
需要算哪些颜色贡献计算一次，哪些计算两次
遍历次数比较多，加读入挂才过
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 7 // fastIO 8 namespace fastIO { 9 #define BUF_SIZE 100000 10 //fread -> read 11 bool IOerror = 0; 12 inline char nc() { 13 static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE; 14 if(p1 == pend) { 15 p1 = buf; 16 pend = buf + fread(buf, 1, BUF_SIZE, stdin); 17 if(pend == p1) { 18 IOerror = 1; 19 return -1; 20 } 21 } 22 return *p1++; 23 } 24 inline bool blank(char ch) { 25 return ch == ' ' || ch == ' ' || ch == ' ' || ch == ' '; 26 } 27 inline void read(int &x) { 28 char ch; 29 while(blank(ch = nc())); 30 if(IOerror) 31 return; 32 for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0'); 33 } 34 #undef BUF_SIZE 35 }; 36 using namespace fastIO; 37 38 typedef long long LL; 39 const int maxn = 2e5 + 10; 40 int vis[maxn], c[maxn]; 41 42 // edge 43 int cnt, h[maxn]; 44 struct edge 45 { 46 int to, pre; 47 } e[maxn<<1]; 48 49 void add(int from, int to) 50 { 51 cnt++; 52 e[cnt].pre = h[from]; 53 e[cnt].to = to; 54 h[from] = cnt; 55 } 56 57 // tree 58 int sz[maxn]; 59 void Size(int x, int f) 60 { 61 sz[x] = 1; 62 int tmp = 0; 63 for(int i = h[x]; i; i = e[i].pre) 64 { 65 int to = e[i].to; 66 if(to == f || vis[to]) continue; 67 Size(to, x); 68 sz[x] += sz[to]; 69 } 70 } 71 72 int rt, M; 73 void Root(int r, int x, int f) 74 { 75 int tmp = 0; 76 for(int i = h[x]; i; i = e[i].pre) 77 { 78 int to = e[i].to; 79 if(to == f || vis[to]) continue; 80 Root(r, to, x); 81 tmp = max(tmp, sz[to]); 82 } 83 tmp = max(tmp, sz[r] - sz[x]); 84 if(tmp < M) M = tmp, rt = x; 85 } 86 87 LL ans; 88 int num[maxn]; 89 void dfs(int x, int rt, int f) 90 { 91 sz[x] = 1; 92 for(int i = h[x]; i; i = e[i].pre) 93 { 94 int to = e[i].to; 95 if(to == f || vis[to]) continue; 96 dfs(to, rt, x); 97 sz[x] += sz[to]; 98 } 99 if(f == rt) ans += (LL) sz[x] * sz[rt]; 100 } 101 102 int c_sz[maxn]; 103 void dfs1(int x, int szr, int f) 104 { 105 num[c[x]]++; 106 if(num[c[x]] == 1) ans += (LL) sz[x] * (szr - c_sz[c[x]]); 107 for(int i = h[x]; i; i = e[i].pre) 108 { 109 int to = e[i].to; 110 if(to == f || vis[to]) continue; 111 dfs1(to, szr, x); 112 } 113 num[c[x]]--; 114 } 115 void dfs2(int x, int f) 116 { 117 num[c[x]]++; 118 if(num[c[x]] == 1) c_sz[c[x]] += sz[x]; 119 for(int i = h[x]; i; i = e[i].pre) 120 { 121 int to = e[i].to; 122 if(to == f || vis[to]) continue; 123 dfs2(to, x); 124 } 125 num[c[x]]--; 126 } 127 void dfs3(int x, int f) 128 { 129 num[c[x]]++; 130 if(num[c[x]] == 1) c_sz[c[x]] -= sz[x]; 131 for(int i = h[x]; i; i = e[i].pre) 132 { 133 int to = e[i].to; 134 if(to == f || vis[to]) continue; 135 dfs3(to, x); 136 } 137 num[c[x]]--; 138 } 139 140 void solve(int x) 141 { 142 M = maxn; 143 Size(x, 0); 144 Root(x, x, 0); 145 vis[rt] = 1; 146 147 dfs(rt, rt, 0); 148 num[c[rt]]++; 149 for(int i = h[rt]; i; i = e[i].pre) 150 { 151 int to = e[i].to; 152 if(vis[to]) continue; 153 dfs1(to, sz[rt] - sz[to], rt); 154 dfs2(to, rt); 155 } 156 for(int i = h[rt]; i; i = e[i].pre) 157 { 158 int to = e[i].to; 159 if(vis[to]) continue; 160 dfs3(to, rt); 161 } 162 num[c[rt]]--; 163 164 for(int i = h[rt]; i; i = e[i].pre) 165 { 166 int to = e[i].to; 167 if(vis[to]) continue; 168 solve(to); 169 } 170 } 171 172 int main(void) 173 { 174 // int size = 256 << 20; // 256MB 175 // char *p = (char*)malloc(size) + size; 176 // __asm__("movl %0, %%esp " :: "r"(p)); 177 178 // freopen("data.txt", "r", stdin); 179 // freopen("ygy.txt", "w", stdout); 180 181 int n, kase = 0; 182 while(read(n), !IOerror) 183 { 184 for(int i = 1; i <= n; i++) read(c[i]); 185 cnt = 0; 186 for(int i = 1; i <= n; i++) h[i] = vis[i] = 0; 187 for(int i = 1; i < n; i++) 188 { 189 int u, v; 190 read(u), read(v); 191 add(u, v), add(v, u); 192 } 193 ans = 0; solve(1); 194 printf("Case #%d: %lld ", ++kase, ans); 195 } 196 return 0; 197 }
1004 Division Game
1005 Expectation Division
1006 Function
观察到置换分解后只能映射到因子长度的环上
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 using namespace std; 6 typedef long long LL; 7 const LL mod = 1e9 + 7; 8 const int maxn = 1e5 + 10; 9 int a[maxn], b[maxn]; 10 int visa[maxn], visb[maxn]; 11 int numa[maxn], numb[maxn]; 12 vector<int> fac[maxn]; 13 14 int main(void) 15 { 16 for(int i = 1; i < maxn; i++) 17 for(int j = i; j < maxn; j += i) 18 fac[j].push_back(i); 19 int n, m, kase= 0; 20 while(~scanf("%d %d", &n, &m)) 21 { 22 for(int i = 0; i < n; i++) scanf("%d", a + i); 23 for(int i = 0; i < m; i++) scanf("%d", b + i); 24 memset(visa, 0, sizeof(visa)); 25 memset(visb, 0, sizeof(visb)); 26 memset(numa, 0, sizeof(numa)); 27 memset(numb, 0, sizeof(numb)); 28 for(int i = 0; i < n; i++) 29 { 30 if(visa[i]) continue; 31 int p = i, cnt = 0; 32 while(!visa[p]) visa[p] = 1, cnt++, p = a[p]; 33 numa[cnt]++; 34 } 35 for(int i = 0; i < m; i++) 36 { 37 if(visb[i]) continue; 38 int p = i, cnt = 0; 39 while(!visb[p]) visb[p] = 1, cnt++, p = b[p]; 40 numb[cnt]++; 41 } 42 LL ans = 1; 43 for(int i = 1; i <= n; i++) 44 { 45 if(!numa[i]) continue; 46 LL tmp = 0; 47 for(int j = 0; j < fac[i].size(); j++) tmp = (tmp + (LL) numb[fac[i][j]] * fac[i][j]) % mod; 48 for(int j = 0; j < numa[i]; j++) ans = ans * tmp % mod; 49 } 50 printf("Case #%d: %lld ", ++kase, ans); 51 } 52 return 0; 53 }
1007 Gear Up
不好写……
1 #include <iostream> 2 #include <cstdio> 3 #include <vector> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7 const int maxn = 1e5 + 10; 8 const int INF = 1e9; 9 vector<int> vec[maxn]; 10 11 int log2(int x) 12 { 13 int cnt = 0; 14 while(x >= 2) x >>= 1, cnt++; 15 return cnt; 16 } 17 18 // edge 19 int cnt, h[maxn]; 20 struct edge 21 { 22 int to, pre; 23 } e[maxn<<1]; 24 void add(int from, int to) 25 { 26 cnt++; 27 e[cnt].pre = h[from]; 28 e[cnt].to = to; 29 h[from] = cnt; 30 } 31 32 // UF 33 int fa[maxn]; 34 int Find(int x) 35 { 36 return fa[x] == x ? x : fa[x] = Find(fa[x]); 37 } 38 void Union(int x, int y) 39 { 40 x = Find(x); 41 y = Find(y); 42 if(x != y) fa[y] = x; 43 } 44 45 // dfs 46 int dfs_clock; 47 int r[maxn], w[maxn]; 48 int vis[maxn], op[maxn], rt[maxn]; 49 int L1[maxn], R1[maxn], L2[maxn], R2[maxn], rl[maxn], rr[maxn]; 50 void dfs(int x, int o, int root) 51 { 52 rt[x] = root; 53 L2[x] = ++dfs_clock; 54 int ww = w[dfs_clock]; 55 vis[x] = 1, op[x] = o; 56 vector<int> tmp = vec[Find(x)]; 57 vec[Find(x)].clear(); 58 int sz = tmp.size(); 59 for(int i = 0; i < sz; i++) 60 { 61 int to = tmp[i]; 62 if(to == x) continue; 63 w[dfs_clock + 1] = ww, dfs(to, 2, root); 64 } 65 R2[x] = dfs_clock, L1[x] = R2[x] + 1; 66 for(int i = h[x]; i; i = e[i].pre) 67 { 68 int to = e[i].to; 69 if(vis[to]) continue; 70 w[dfs_clock + 1] = ww + r[x] - r[to], dfs(to, 1, root); 71 } 72 R1[x] = dfs_clock; 73 } 74 75 // seg 76 int M[maxn<<2], tag[maxn<<2]; 77 void gather(int p) 78 { 79 M[p] = max(M[p<<1], M[p<<1|1]); 80 } 81 void push(int p) 82 { 83 if(tag[p]) 84 { 85 tag[p<<1] += tag[p]; 86 tag[p<<1|1] += tag[p]; 87 M[p<<1] += tag[p]; 88 M[p<<1|1] += tag[p]; 89 tag[p] = 0; 90 } 91 } 92 void build(int p, int l, int r) 93 { 94 tag[p] = 0; 95 if(l < r) 96 { 97 int mid = (l + r) >> 1; 98 build(p<<1, l, mid); 99 build(p<<1|1, mid + 1, r); 100 gather(p); 101 } 102 else M[p] = w[l]; 103 } 104 void modify(int p, int tl, int tr, int l, int r, int v) 105 { 106 if(tl > tr) return; 107 if(tr < l || r < tl) return; 108 if(l <= tl && tr <= r) 109 { 110 tag[p] += v; 111 M[p] += v; 112 return; 113 } 114 push(p); 115 int mid = (tl + tr) >> 1; 116 modify(p<<1, tl, mid, l, r, v); 117 modify(p<<1|1, mid+1, tr, l, r, v); 118 gather(p); 119 } 120 int query(int p, int tl, int tr, int l, int r) 121 { 122 if(tl > tr) return -INF; 123 if(tr < l || r < tl) return -INF; 124 if(l <= tl && tr <= r) return M[p]; 125 push(p); 126 int mid = (tl + tr) >> 1; 127 return max(query(p<<1, tl, mid, l, r), query(p<<1|1, mid+1, tr, l, r)); 128 } 129 130 int main(void) 131 { 132 int n, m, q, kase = 0; 133 while(~scanf("%d %d %d", &n, &m, &q)) 134 { 135 for(int i = 1; i <= n; i++) 136 { 137 int x; 138 scanf("%d", &x); 139 r[i] = log2(x); 140 } 141 cnt = dfs_clock = 0; 142 for(int i = 1; i <= n; i++) vis[i] = h[i] = 0, fa[i] = i; 143 for(int i = 1; i <= m; i++) 144 { 145 int a, x, y; 146 scanf("%d %d %d", &a, &x, &y); 147 if(a == 1) add(x, y), add(y, x); 148 else Union(x, y); 149 } 150 for(int i = 1; i <= n; i++) vec[Find(i)].push_back(i); 151 for(int i = 1; i <= n; i++) 152 if(!vis[i]) w[dfs_clock + 1] = 0, rl[i] = dfs_clock + 1, dfs(i, 0, i), rr[i] = dfs_clock; 153 build(1, 1, n); 154 printf("Case #%d: ", ++kase); 155 while(q--) 156 { 157 int a, x, y; 158 scanf("%d %d %d", &a, &x, &y), y = log2(y); 159 if(a == 1) 160 { 161 if(op[x] % 2) modify(1, 1, n, L2[x], R2[x], r[x] - y); 162 else modify(1, 1, n, L1[x], R1[x], y - r[x]); 163 r[x] = y; 164 } 165 else printf("%.3f ", log(2) * (y + query(1, 1, n, rl[rt[x]], rr[rt[x]]) - query(1, 1, n, L2[x], L2[x]))); 166 } 167 } 168 return 0; 169 }
1008 Hints of sd0061
NthElement 实现上是类似快排 由于只递归一边所以期望on
按询问从大到小做 不同的询问不多 每次排的长度递减 据说就on了
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <algorithm> 5 using namespace std; 6 7 unsigned a[10000010]; 8 // Nth_element 9 int Partition(unsigned a[], int l, int r) 10 { 11 int rd = l + (long long) rand() * 10007 % (r - l + 1); 12 swap(a[rd], a[r]); 13 int j = l - 1; 14 for(int i = l; i < r; i++) if(a[i] <= a[r]) swap(a[++j], a[i]); 15 swap(a[++j], a[r]); 16 return j; 17 } 18 int NthElement(unsigned a[], int l, int r, int id) 19 { 20 if(l == r) return a[l]; 21 int m = Partition(a, l, r), cur = m - l + 1; 22 if(id == cur) return a[m]; 23 if(id < cur) return NthElement(a, l, m - 1, id); 24 return NthElement(a, m + 1, r, id - cur); 25 } 26 27 // rand 28 unsigned x, y, z; 29 unsigned rng61() { 30 unsigned t; 31 x ^= x << 16; 32 x ^= x >> 5; 33 x ^= x << 1; 34 t = x; 35 x = y; 36 y = z; 37 z = t ^ x ^ y; 38 return z; 39 } 40 41 int q[222], p[222], ans[222]; 42 bool cmp(int i, int j) 43 { 44 return q[i] > q[j]; 45 } 46 47 int main(void) 48 { 49 unsigned A, B, C; 50 int n, m, kase = 0; 51 while(~scanf("%d %d %u %u %u", &n, &m, &A, &B, &C)) 52 { 53 for(int i = 1; i <= m; i++) scanf("%d", q + i), p[i] = i; 54 x = A, y = B, z = C; 55 for(int i = 1; i <= n; i++) a[i] = rng61(); 56 sort(p + 1, p + 1 + m, cmp); 57 int mi = n; 58 for(int i = 1; i <= m; i++) 59 { 60 if(i > 1 && q[p[i]] == q[p[i-1]]) ans[p[i]] = ans[p[i-1]]; 61 else 62 { 63 ans[p[i]] = NthElement(a, 1, mi, q[p[i]] + 1); 64 mi = q[p[i]] + 1; 65 } 66 } 67 printf("Case #%d:", ++kase); 68 for(int i = 1; i <= m; i++) printf(" %u", ans[i]); puts(""); 69 } 70 return 0; 71 }
1009 I Curse Myself
仙人掌环抠出来 k路归并
1 #include <iostream> 2 #include <cstdio> 3 #include <vector> 4 #include <algorithm> 5 using namespace std; 6 typedef long long LL; 7 typedef pair<int, int> pii; 8 typedef pair<int, pii> piii; 9 int k; 10 11 // edge 12 int cnt, h[1111]; 13 struct edge 14 { 15 int to, pre, v; 16 } e[4444]; 17 void add(int from, int to, int v) 18 { 19 cnt++; 20 e[cnt].pre = h[from]; 21 e[cnt].to = to; 22 e[cnt].v = v; 23 h[from] = cnt; 24 } 25 26 bool cmp(int i, int j) 27 { 28 return i > j; 29 } 30 31 // loop 32 int vis[1111], fa[1111], fae[1111], num; 33 vector<int> seq[1111]; 34 void dfs(int x, int f, int eid) 35 { 36 vis[x] = 1, fa[x] = f; fae[x] = eid; 37 for(int i = h[x]; i; i = e[i].pre) 38 { 39 int to = e[i].to; 40 if(to == f || vis[to] == 2) continue; 41 if(vis[to] == 1) 42 { 43 seq[++num].clear(); 44 seq[num].push_back(e[i].v); 45 int tmp = x; 46 while(tmp != to) 47 { 48 seq[num].push_back(e[fae[tmp]].v); 49 tmp = fa[tmp]; 50 } 51 sort(seq[num].begin(), seq[num].end(), cmp); 52 } 53 else dfs(to, x, i); 54 } 55 vis[x] = 2; 56 } 57 58 vector<int> ret; 59 60 struct PQ 61 { 62 piii a[222222]; 63 int tail; 64 PQ(){tail = 0;} 65 void push(piii cur) 66 { 67 a[++tail] = cur; 68 int p = tail; 69 while(p > 1) 70 { 71 if(a[p] > a[p/2]) swap(a[p], a[p/2]), p /= 2; 72 else break; 73 } 74 } 75 piii top() 76 { 77 return a[1]; 78 } 79 void pop() 80 { 81 a[1] = a[tail--]; 82 int p = 1; 83 while(p <= tail) 84 { 85 if(p + p <= tail && a[p] < a[p+p] && (p + p + 1 > tail || a[p+p] > a[p+p+1])) swap(a[p], a[p+p]), p *= 2; 86 else if(p + p + 1 <= tail && a[p] < a[p+p+1]) swap(a[p], a[p+p+1]), p = p + p + 1; 87 else break; 88 } 89 } 90 bool empty() 91 { 92 return tail == 0; 93 } 94 } pq; 95 96 void Merge(vector<int> & ans, vector<int> & nxt) 97 { 98 int sza = ans.size(), szn = nxt.size(); 99 pq.tail = 0; 100 ret.resize(0); 101 for(int i = 0; i < szn; i++) pq.push(piii(ans[0] + nxt[i], pii(0, i))); 102 while(ret.size() < k && !pq.empty()) 103 { 104 piii tmp = pq.top(); pq.pop(); 105 ret.push_back(tmp.first); 106 if(tmp.second.first + 1 < sza) pq.push(piii(ans[tmp.second.first+1] + nxt[tmp.second.second], pii(tmp.second.first + 1, tmp.second.second))); 107 } 108 ans = ret; 109 } 110 111 int main(void) 112 { 113 // freopen("1009.in", "r", stdin); 114 // freopen("ygy.txt", "w", stdout); 115 116 int n, m, kase = 0; 117 while(~scanf("%d %d", &n, &m)) 118 { 119 int sum = cnt = 0; 120 for(int i = 1; i <= n; i++) h[i] = 0, vis[i] = 0; 121 for(int i = 1; i <= m; i++) 122 { 123 int x, y, z; 124 scanf("%d %d %d", &x, &y, &z); 125 sum += z, add(x, y, z), add(y, x, z); 126 } 127 scanf("%d", &k); 128 129 num = 0, dfs(1, 0, 0); 130 vector<int> ans = seq[1]; 131 for(int i = 2; i <= num; i++) Merge(ans, seq[i]); 132 133 unsigned int ret = 0, sza = ans.size(); 134 if(!num) ret += sum; 135 else for(int i = 0; i < sza; i++) ret += (i + 1) * (sum - ans[i]); 136 printf("Case #%d: %u ", ++kase, ret); 137 } 138 return 0; 139 }
1011 KazaQ's Socks
找找规律。
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 typedef long long LL; 5 6 int main(void) 7 { 8 int kase = 0; 9 LL n, k; 10 while(~scanf("%lld %lld", &n, &k)) 11 { 12 LL q = (k - 1) / (n - 1), r = (k - 1) % (n - 1) + 1, ans; 13 if(q == 0) ans = r; 14 else if(q % 2) ans = r == 1 ? n : r - 1; 15 else ans = r == 1 ? n - 1 : r - 1; 16 printf("Case #%d: %lld ", ++kase, ans); 17 } 18 return 0; 19 }
1012 Limited Permutation
加读入挂直接log就过了
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <vector> 5 using namespace std; 6 7 // fastIO 8 namespace fastIO { 9 #define BUF_SIZE 100000 10 //fread -> read 11 bool IOerror = 0; 12 inline char nc() { 13 static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE; 14 if(p1 == pend) { 15 p1 = buf; 16 pend = buf + fread(buf, 1, BUF_SIZE, stdin); 17 if(pend == p1) { 18 IOerror = 1; 19 return -1; 20 } 21 } 22 return *p1++; 23 } 24 inline bool blank(char ch) { 25 return ch == ' ' || ch == ' ' || ch == ' ' || ch == ' '; 26 } 27 inline void read(int &x) { 28 char ch; 29 while(blank(ch = nc())); 30 if(IOerror) 31 return; 32 for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0'); 33 } 34 #undef BUF_SIZE 35 }; 36 using namespace fastIO; 37 38 typedef long long LL; 39 typedef pair<int, int> pii; 40 const LL mod = 1e9 + 7; 41 const int maxn = 1e6 + 10; 42 int l[maxn], r[maxn]; 43 vector<pii> v[maxn]; 44 LL inv[maxn], fac[maxn], ifac[maxn]; 45 46 LL C(int a, int b) 47 { 48 return fac[a] * ifac[b] % mod * ifac[a-b] % mod; 49 } 50 51 LL solve(int L, int R) 52 { 53 if(L > R) return 1LL; 54 pii now = v[L].back(); v[L].pop_back(); 55 if(now.first != R) return 0LL; 56 return solve(L, now.second - 1) * solve(now.second + 1, R) % mod * C(R - L, now.second - L) % mod; 57 } 58 59 int main(void) 60 { 61 fac[0] = ifac[0] = inv[1] = 1; 62 for(int i = 2; i < maxn; i++) inv[i] = (mod - (mod / i) * inv[mod%i] % mod) % mod; 63 for(int i = 1; i < maxn; i++) fac[i] = fac[i-1] * i % mod, ifac[i] = ifac[i-1] * inv[i] % mod; 64 int n, kase = 0; 65 while(read(n), !IOerror) 66 { 67 for(int i = 1; i <= n; i++) read(l[i]), v[i].clear(); 68 for(int i = 1; i <= n; i++) read(r[i]), v[l[i]].push_back(pii(r[i], i)); 69 for(int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end()); 70 LL ans = solve(1, n); 71 printf("Case #%d: %lld ", ++kase, ans); 72 } 73 return 0; 74 }