Day1:
第一题：水题
#include <iostream> #include <cstdio> #include <cstring> using namespace std; inline long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) f *= (ch == '-') ? -1 : 1, ch = getchar(); do ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar(); while(isdigit(ch)); return ans * f; } const int MAXN = 1005; char s1[MAXN], s2[MAXN]; inline int ord(char c){ if(c >= 'a' && c <= 'z') return c - 'a'; return c - 'A'; } int main(){ scanf("%s%s", s1, s2); int len1 = strlen(s1), len2 = strlen(s2); for(int i=0, j=0; i<len2; i++, j=(j+1)%len1) putchar((26 + ord(s2[i]) - ord(s1[j])) % 26 + ((s2[i] >= 'a') ? 'a' : 'A')); return 0; }
第二题：贪心提前处理出最后的序列（以$a*b$为关键字排序），最后高精度模拟即可
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; inline long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) f *= (ch == '-') ? -1 : 1, ch = getchar(); do ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar(); while(isdigit(ch)); return ans * f; } const int MAXN = 1005; struct Node{ int a, b; const bool operator < (const Node &x) const{ return a * b < x.a * x.b; } }a[MAXN]; const int BITLEN = 10000, BIGSIZE = 900; struct Big{ int val[BIGSIZE+1]; void operator = (int x){ memset(val, 0, sizeof(val)); val[0] = x; } Big(int x = 0){this->operator = (x);} Big operator + (Big x){ Big ans; for(int i=0; i<BIGSIZE; i++){ ans.val[i] += val[i] + x.val[i]; ans.val[i + 1] += ans.val[i] / BITLEN; ans.val[i] %= BITLEN; } return ans; } Big operator * (int x){ Big ans; for(int i=0; i<BIGSIZE; i++) ans.val[i] = x * val[i]; for(int i=0; i<BIGSIZE-1; i++) ans.val[i+1] += ans.val[i] / BITLEN, ans.val[i] %= BITLEN; return ans; } Big operator / (int x){ Big ans; for(int i=BIGSIZE-1, rest=0; i>=0; i--){ rest = rest * BITLEN + val[i]; ans.val[i] = rest / x; rest = rest % x; } return ans; } const bool operator < (const Big &x)const{ for(int i=BIGSIZE-1; i>=0; i--) if(val[i] != x.val[i]) return val[i] < x.val[i]; return false; } void print(){ int len = BIGSIZE - 1; while(!val[len]) len--; for(int i=len; i>=0; i--){ if(i != len) for(int j=BITLEN/10; j>1; j/=10) if(val[i] < j) putchar('0'); printf("%d", val[i]); } } }; int main(){ int n = read(); for(int i=0; i<=n; i++) a[i].a = read(), a[i].b = read(); sort(a+1, a+1+n); Big ans = 0, sum = a[0].a; for(int i=1; i<=n; i++){ Big temp = sum / a[i].b; if(ans < temp) ans = temp; sum = sum * a[i].a; } ans.print(); return 0; }
第三题：倍增Dp提前预处理出一个点走$2^i$后的情况，对于寻找第一近和第二近的房子可以用map进行维护，每次找到最近的几个点，最后直接模拟即可
#include <iostream> #include <cstdio> #include <cstring> #include <map> #include <cmath> #include <queue> using namespace std; inline long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) f *= (ch == '-') ? -1 : 1, ch = getchar(); do ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar(); while(isdigit(ch)); return ans * f; } const int MAXN = 100001, LOGN = 18; int h[MAXN], next1[MAXN], next2[MAXN]; long long Dp[MAXN][LOGN+1][2], last[MAXN][LOGN+1]; map<int,int> hash; struct Tuple{ int dis, h, id; const bool operator < (const Tuple &x) const{ if(dis != x.dis) return dis > x.dis; return h > x.h; } }; priority_queue<Tuple> Q; int main(){ int n = read(); for(int i=1; i<=n; i++) h[i] = read(); hash[h[n]] = n, hash[h[n-1]] = n - 1; next1[n] = next2[n] = n, next1[n-1] = n, next2[n-1] = n-1; for(int i=n-2; i>=1; i--){ map<int,int>::iterator it = hash.upper_bound(h[i]), it2 = it; for(int j=1; j<=2; j++, it++) if(it != hash.end()) Q.push((Tuple){abs(h[i]-it->first), it->first, it->second}); if(it2 != hash.begin()){ it2--; for(int j=1; j<=2; j++, it2--){ Q.push((Tuple){abs(h[i]-it2->first), it2->first, it2->second}); if(it2 == hash.begin()) break; } } next1[i] = Q.top().id; Q.pop(); next2[i] = Q.top().id; hash[h[i]] = i; while(!Q.empty()) Q.pop(); } for(int i=1; i<=n; i++) Dp[i][0][0] = abs(h[next2[i]] - h[i]), Dp[i][0][1] = abs(h[next1[i]] - h[i]); for(int i=1; i<=n; i++) Dp[i][1][0] = Dp[i][0][0], Dp[i][1][1] = (next2[i] == i) ? 0 : Dp[next2[i]][0][1], last[i][1] = (next2[i] == i) ? next2[i] : next1[next2[i]]; for(int i=2; i<=LOGN; i++) for(int j=1; j<=n; j++){ Dp[j][i][0] = Dp[last[j][i-1]][i-1][0] + Dp[j][i-1][0], Dp[j][i][1] = Dp[last[j][i-1]][i-1][1] + Dp[j][i-1][1]; last[j][i] = last[last[j][i-1]][i-1]; } int x0 = read(), s0 = 0; double maxNum = 1e18; for(int i=1; i<=n; i++){ long long disA = 0, disB = 0, u = i, x = x0; for(int j=LOGN; j>=1; j--) if(Dp[u][j][0] + Dp[u][j][1] <= x) disA += Dp[u][j][0], disB += Dp[u][j][1], x -= Dp[u][j][0] + Dp[u][j][1], u = last[u][j]; if(Dp[u][0][0] <= x) disA += Dp[u][0][0]; if((double)disA / disB < maxNum) maxNum = (double)disA / disB, s0 = i; } printf("%d ", s0); int m = read(); for(int i=1; i<=m; i++){ long long disA = 0, disB = 0, u = read(), x = read(); for(int j=LOGN; j>=1; j--) if(Dp[u][j][0] + Dp[u][j][1] <= x) disA += Dp[u][j][0], disB += Dp[u][j][1], x -= Dp[u][j][0] + Dp[u][j][1], u = last[u][j]; if(Dp[u][0][0] <= x) disA += Dp[u][0][0]; printf("%lld %lld ", disA, disB); } return 0; }
第四题：对于$n<=1000$，暴力即可，对于$n<=10^6，z=0$，使用树状数组优化即可，对于数据加强的$11,12$点，可以考虑用树套树，但也可以用CDQ
因为CDQ的前提必须是更新过的节点不再更新，所以先分治，后计算左对右贡献的做法显然不行，但是可以先分治左，再计算左对右，再计算右，需要时刻牢记自己定义的CDQ是什么
这道题CDQ函数定义：将L到R的已经对X排序的数组计算贡献，并最后按照Y进行排序。
这样如果数组顺序不对需要重新排序（用sort排序），当然因为sort是在第一层CDQ里面的所以总时间复杂度还是$O(nlog^2n)$，和树套树的时间复杂度一样（虽然常数大了一点）
普通代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; inline long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) f *= (ch == '-') ? -1 : 1, ch = getchar(); do ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar(); while(isdigit(ch)); return ans * f; } const int MAXN = 1000001; struct Tuple{ int x, y, z, val; bool const operator < (const Tuple &temp) const{ if(x != temp.x) return x < temp.x; if(y != temp.y) return y < temp.y; return z < temp.z; } }tup[MAXN]; struct BIT{ int val[MAXN]; void add(int pos,int x){ for(int i=pos; i<MAXN; i+=i&(-i)) val[i] = max(val[i], x); } int query(int pos){ int ans = 0; for(int i=pos; i; i-=i&(-i)) ans = max(ans, val[i]); return ans; } }bit; int main(){ int n = read(); for(int i=1; i<=n; i++) tup[i].x = read(), tup[i].y = read(), tup[i].z = read(), tup[i].val = 1; sort(tup+1, tup+1+n); if(n > 10001){ for(int i=1; i<=n; i++){ tup[i].val = bit.query(tup[i].y) + 1; bit.add(tup[i].y, tup[i].val); } } else{ for(int i=1; i<=n; i++) for(int j=1; j<i; j++) if(tup[j].y <= tup[i].y && tup[j].z <= tup[i].z) tup[i].val = max(tup[i].val, tup[j].val + 1); } int ans = 0; for(int i=1; i<=n; i++) ans = max(ans, tup[i].val); printf("%d", ans); return 0; }
数据加强版代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; inline long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) f *= (ch == '-') ? -1 : 1, ch = getchar(); do ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar(); while(isdigit(ch)); return ans * f; } const int MAXN = 1000005; struct Tuple{ int x, y, z, val; bool const operator < (const Tuple &temp) const{ if(x != temp.x) return x < temp.x; if(y != temp.y) return y < temp.y; return z < temp.z; } }tup[MAXN], temp[MAXN]; struct BIT{ int val[MAXN]; void add(int pos,int x){ for(int i=pos; i<MAXN; i+=i&(-i)) val[i] = max(val[i], x); } int query(int pos){ int ans = 0; for(int i=pos; i; i-=i&(-i)) ans = max(ans, val[i]); return ans; } void clear(int pos){ for(int i=pos; i<MAXN; i+=i&(-i)) val[i] = 0; } }bit; const bool cmp(const Tuple &a,const Tuple &b){ if(a.y != b.y) return a.y < b.y; return a.z < b.z; } void CDQ(int l,int r){ int mid = (l + r) >> 1; if(l == r) return; CDQ(l, mid); sort(tup+mid+1, tup+r+1, cmp); for(int i=l, j=mid+1, k=l; k<=r; k++){ if(j > r || (i <= mid && tup[i].y <= tup[j].y)) bit.add(tup[i].z, tup[i].val), i++; else tup[j].val = max(tup[j].val, bit.query(tup[j].z) + 1), j++; } for(int i=l; i<=mid; i++) bit.clear(tup[i].z); sort(tup+mid+1, tup+r+1); CDQ(mid+1, r); for(int i=l, j=mid+1, k=l; k<=r; k++){ if(j > r || (i <= mid && tup[i].y <= tup[j].y)) temp[k] = tup[i++]; else temp[k] = tup[j++]; } for(int i=l; i<=r; i++) tup[i] = temp[i]; } int main(){ int n = read(); for(int i=1; i<=n; i++) tup[i].x = read() + 1, tup[i].y = read() + 1, tup[i].z = read() + 1, tup[i].val = 1; sort(tup+1, tup+1+n); CDQ(1, n); int ans = 0; for(int i=1; i<=n; i++) ans = max(ans, tup[i].val); printf("%d", ans); return 0; }
Day2:
第一题：数论水题，恰好用上了我改进版的扩展欧几里得函数
第二题：线段树区间修改，区间查询的题目，当然也可以用前缀数组加上二分答案来做
线段树代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; inline long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) f *= (ch == '-') ? -1 : 1, ch = getchar(); do ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar(); while(isdigit(ch)); return ans * f; } const int MAXN = 1000005; int val[MAXN]; struct Segment{ struct Node{ int l, r, isSigned, sign, minNum; }node[MAXN*4]; inline void mix(int x,int sign){ node[x].sign += sign; node[x].minNum += sign; node[x].isSigned = true; } void spread(int x){ if(node[x].isSigned){ mix(x*2, node[x].sign); mix(x*2+1, node[x].sign); node[x].sign = 0; node[x].isSigned = false; } } bool change(int l,int r,int sign,int x = 1){ if(l <= node[x].l && node[x].r <= r){ mix(x, sign); return node[x].minNum >= 0; } spread(x); bool flag = true; if(node[x*2].r >= l) flag &= change(l, r, sign, x*2); if(node[x*2+1].l <= r) flag &= change(l, r, sign, x*2+1); node[x].minNum = min(node[x*2].minNum, node[x*2+1].minNum); return flag; } void build(int l,int r,int x = 1){ node[x].l = l, node[x].r = r; if(l == r){ node[x].minNum = node[x].sign = val[l]; return; } int mid = (l + r) >> 1; build(l, mid, x*2); build(mid+1, r, x*2+1); node[x].minNum = min(node[x*2].minNum, node[x*2+1].minNum); } }T; int main(){ int n = read(), m = read(); for(int i=1; i<=n; i++) val[i] = read(); T.build(1, n); for(int i=1; i<=m; i++){ int v = read(), l = read(), r = read(); if(!T.change(l, r, -v)){ printf("-1 %d", i); return 0; } } printf("0"); return 0; }
前缀数组二分答案做法：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; inline long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) f *= (ch == '-') ? -1 : 1, ch = getchar(); do ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar(); while(isdigit(ch)); return ans * f; } const int MAXN = 1000005; long long val[MAXN], temp[MAXN]; int Qv[MAXN], Ql[MAXN], Qr[MAXN]; int main(){ int n = read(), m = read(); for(int i=1; i<=n; i++) val[i] = read(); for(int i=1; i<=m; i++) Qv[i] = read(), Ql[i] = read(), Qr[i] = read(); int l = 1, r = m + 1; while(l < r){ int mid = (l + r) >> 1; for(int i=1; i<=n; i++) temp[i] = val[i]; for(int i=n; i>=2; i--) temp[i] -= temp[i-1]; for(int i=1; i<=mid; i++) temp[Ql[i]] += -Qv[i], temp[Qr[i] + 1] -= -Qv[i]; for(int i=2; i<=n; i++) temp[i] += temp[i-1]; bool flag = true; for(int i=1; i<=n; i++) flag &= (temp[i] >= 0); if(flag) l = mid + 1; else r = mid; } if(l == m + 1) printf("0"); else printf("-1 %d", l); return 0; }
第三题：图论题，首先使用倍增优化优化提点的过程，之后考虑贪心，二分答案，对答案$mid$
将所有的军队分为三类：
1.不能走到1节点，那就尽可能向上走
2.能走到1节点但不能走回来，这样如果这个子节点需要军队驻扎就驻扎在这里，否则就走到1节点
3.能走到1节点，还能走回来，这种军队就走到1节点
最后再对没走到的子节点进行分配即可
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; inline long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) f *= (ch == '-') ? -1 : 1, ch = getchar(); do ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar(); while(isdigit(ch)); return ans * f; } const int MAXN = 50001, LOGN = 16; int head[MAXN], next[MAXN<<1], last[MAXN<<1], val[MAXN<<1], lineNum = 0; void add(int x,int y,int v){ lineNum++, next[lineNum] = head[x], last[lineNum] = y, val[lineNum] = v, head[x] = lineNum; } long long Dp[MAXN][LOGN+1], fa[MAXN][LOGN+1]; int pos[MAXN], n, m, cnt = 0; pair<long long,int> temp[MAXN], need[MAXN]; bool isRest[MAXN], isNeed[MAXN]; void DFS(int x,int pre){ for(int i=1; i<=LOGN; i++) Dp[x][i] = Dp[fa[x][i-1]][i-1] + Dp[x][i-1], fa[x][i] = fa[fa[x][i-1]][i-1]; for(int l=head[x]; l; l=next[l]){ int y = last[l]; if(y != pre) Dp[y][0] = val[l], fa[y][0] = x, DFS(y, x); } } bool BFS(int x,int pre){ bool isSon = true, flag = false; for(int l=head[x]; l; l=next[l]){ int y = last[l]; if(y == pre) continue; isSon = false, BFS(y, x); flag |= isNeed[y]; } if(!isSon) isNeed[x] &= flag; } bool check(long long mid){ memset(isNeed, true, sizeof(isNeed)); memset(isRest, false, sizeof(isRest)); for(int i=1; i<=m; i++) temp[i] = make_pair(mid, pos[i]); for(int i=1; i<=m; i++) for(int j=LOGN; j>=0; j--) if(fa[temp[i].second][j] != 1 && temp[i].first >= Dp[temp[i].second][j]) temp[i].first -= Dp[temp[i].second][j], temp[i].second = fa[temp[i].second][j]; int cntRest = m, cntLast = 0; for(int i=1; i<=m; i++) if(temp[i].first < Dp[temp[i].second][0]) temp[i].first = 1e18, cntRest--, isNeed[temp[i].second] = false; for(int i=1; i<=cnt; i++) BFS(need[i].second, 1); cntLast = cntRest; sort(temp+1, temp+1+m); for(int i=1; i<=cntRest; i++) if(temp[i].first < 2 * Dp[temp[i].second][0] && isNeed[temp[i].second]) temp[i].first = 1e18, cntLast--, isNeed[temp[i].second] = false; else temp[i].first -= Dp[temp[i].second][0]; sort(temp+1, temp+1+cntRest); cntRest = cntLast; int i = 1, j = 1; while(1){ while(!isNeed[need[j].second] && j<=cnt) j++; if(j > cnt) return true; if(i > cntRest) return false; if(temp[i].first < need[j].first) i++; else i++, j++; } } int main(){ n = read(); long long l = 0, r = 0; for(int i=1; i<n; i++){ int x = read(), y = read(), v = read(); add(x, y, v), add(y, x, v), r += v; } for(int i=0; i<=LOGN; i++) fa[1][i] = 1; DFS(1, 1); for(int i=2; i<=n; i++) if(fa[i][0] == 1) need[++cnt] = make_pair(Dp[i][0], i); sort(need+1, need+1+cnt); m = read(); for(int i=1; i<=m; i++) pos[i] = read(); bool flag = false; while(l < r){ int mid = (l + r) >> 1; if(check(mid)) r = mid, flag = true; else l = mid + 1; } if(flag) printf("%d", l); else printf("-1"); return 0; }