A：水题
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> pii; typedef tuple<int,int,int> tu; const int N = 1e6 + 5; const int M = 1e3 + 5; const double eps = 1e-10; const LL Mod = 1e9 + 7; #define pi acos(-1) #define INF 1e9 #define dbg(ax) cout << "now this num is " << ax << endl; inline int read() { int f = 1;int x = 0;char c = getchar(); while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } inline long long ADD(long long x,long long y) {return (x + y) % Mod;} inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;} inline long long MUL(long long x,long long y) {return x * y % Mod;} void solve() { int n;n = read(); string s;cin >> s; string ans = ""; for(int i = 0;i < s.size();++i) { if(s[i] == 'U') ans += "D"; else if(s[i] == 'D') ans += 'U'; else if(s[i] == 'L') ans += "LR"; } cout << ans << " "; } int main() { int ca;ca = read(); while(ca--) { solve(); } //system("pause"); return 0; }
B：维护一下二进制位即可。
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> pii; typedef tuple<int,int,int> tu; const int N = 3e5 + 5; const int M = 1e3 + 5; const double eps = 1e-10; const LL Mod = 1e9 + 7; #define pi acos(-1) #define INF 1e9 #define dbg(ax) cout << "now this num is " << ax << endl; inline int read() { int f = 1;int x = 0;char c = getchar(); while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } inline long long ADD(long long x,long long y) {return (x + y) % Mod;} inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;} inline long long MUL(long long x,long long y) {return x * y % Mod;} int pre[N]; void init() { pre[0] = 0; for(int i = 1;i < N;++i) pre[i] = pre[i - 1] ^ i; } void solve() { int a,b;a = read(),b = read(); int ma = pre[a - 1],ans = a; if(ma == b) printf("%d ",ans); else { int tmp = 0; for(int i = 29;i >= 0;--i) { int g = (ma >> i) & 1; int g2 = (b >> i) & 1; if(g ^ g2 == 1) tmp += (1 << i); } if(tmp != a) printf("%d ",ans + 1); else printf("%d ",ans + 2); } } int main() { init(); int ca;ca = read(); while(ca--) { solve(); } //system("pause"); return 0; }
C：这题质量其实挺高的。
因为第i位如果不从当前位相加得到，那么最多可以从i - 2的位置借1个值。
那么可以发现状态并不是很多。
记忆化dfs即可。
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> pii; typedef tuple<int,int,int> tu; const int N = 3e5 + 5; const int M = 1e3 + 5; const double eps = 1e-10; const LL Mod = 1e9 + 7; #define pi acos(-1) #define INF 1e9 #define dbg(ax) cout << "now this num is " << ax << endl; inline int read() { int f = 1;int x = 0;char c = getchar(); while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } inline long long ADD(long long x,long long y) {return (x + y) % Mod;} inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;} inline long long MUL(long long x,long long y) {return x * y % Mod;} LL dp[15][2][2][2][2]; string s; bool check(int x,int y) { if(x <= 8 && y == x + 1) return true; else return false; } LL dfs(int len,bool ned1,bool ned2,bool limit1,bool limit2) { if(len == s.size()) { if(limit2 || limit1) return 0; if(!ned1 && !ned2) return 1; else return 0; } if(dp[len][ned1][ned2][limit1][limit2] != -1) return dp[len][ned1][ned2][limit1][limit2]; int now = s[len] - '0'; LL ans = 0; for(int i = 0;i <= 9;++i) { for(int j = 0;j <= 9;++j) { int x = i + j,y = (i + j) % 10; if(ned1) { if(x >= 10) { if(y == now) ans += dfs(len + 1,ned2,false,limit1 & (i == 0),limit2 & (j == 0)); else if(check(y,now)) ans += dfs(len + 1,ned2,true,limit1 & (i == 0),limit2 & (j == 0)); } else if(x == 9 && now == 0) ans += dfs(len + 1,ned2,true,limit1 & (i == 0),limit2 & (j == 0)); } else { if(x <= 8) { if(y == now) ans += dfs(len + 1,ned2,false,limit1 & (i == 0),limit2 & (j == 0)); else if(y == now - 1) ans += dfs(len + 1,ned2,true,limit1 & (i == 0),limit2 & (j == 0)); } else if(x == 9 && y == now) ans += dfs(len + 1,ned2,false,limit1 & (i == 0),limit2 & (j == 0)); } } } // if(ans != 0) printf("len is %d ned1 is %d ned2 is %d ans is %d ",len,ned1,ned2,ans); return dp[len][ned1][ned2][limit1][limit2] = ans; } void solve() { memset(dp,-1,sizeof(dp)); cin >> s; LL ans = dfs(0,false,false,true,true); printf("%lld ",ans); } int main() { int ca;ca = read(); while(ca--) { solve(); } // system("pause"); return 0; }
D：比C题还水感觉。
显然要让答案最大，我们要维护的最高位出现的次数尽可能多。
所以考虑从最高位开始取。这个对于取的循环，从小到大还是从大到小会不一样。
我们考虑从小到大，因为这样能保证后面的最高位也尽可能多。
即对于100 3 得80 + 10 + 10//从小到大、
得90 5 5.//从大到小。
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> pii; typedef tuple<int,int,int> tu; const int N = 1e6 + 5; const int M = 2e4 + 5; const double eps = 1e-10; const LL Mod = 1e9 + 7; #define pi acos(-1) #define INF 1e9 #define dbg(ax) cout << "now this num is " << ax << endl; inline int read() { int f = 1;int x = 0;char c = getchar(); while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } inline long long ADD(long long x,long long y) {return (x + y) % Mod;} inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;} inline long long MUL(long long x,long long y) {return x * y % Mod;} int a[15],pw[15],ans[105]; void solve() { pw[0] = 1; for(int i = 1;i <= 9;++i) pw[i] = pw[i - 1] * 10; int ca;ca = read(); while(ca--) { int s,n;s = read(),n = read(); memset(ans,0,sizeof(ans)); for(int i = 1;i <= n;++i) { int len = 0; int x = s; while(x) { a[++len] = x % 10; x /= 10; } int g = n - i; for(int k = len;k >= 1;--k) { if(ans[i] != 0) break; for(int j = 1;j <= 9;++j) { LL ss = s - 1LL * j * pw[k - 1]; if(ss >= g) { // printf("ss is %d g is %d pw[%d] is %d ",ss,g,k - 1,pw[k - 1]); ans[i] = j * pw[k - 1]; s -= j * pw[k - 1]; break; } } } } if(s != 0) ans[1] += s; for(int i = 1;i <= n;++i) printf("%d%c",ans[i],i == n ? ' ' : ' '); } } int main() { solve(); //system("pause"); return 0; }
E：这题感觉也挺简单的。
线段树上维护一下左右的最大长度和左右的值即可。
能连起来就合并一下答案。
有一个细节就是query的时候可能有需要合并的答案需要我们手动去合并。
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> pii; typedef tuple<int,int,int> tu; const int N = 2e5 + 5; const int M = 2e4 + 5; const double eps = 1e-10; const LL Mod = 1e9 + 7; #define pi acos(-1) #define INF 1e9 #define dbg(ax) cout << "now this num is " << ax << endl; inline int read() { int f = 1;int x = 0;char c = getchar(); while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } inline long long ADD(long long x,long long y) {return (x + y) % Mod;} inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;} inline long long MUL(long long x,long long y) {return x * y % Mod;} int a[N]; struct Node{int L,r,Lmx,rmx,Lnum,rnum;LL sum;}node[N << 2]; LL cal(int x) { return 1LL * x * (x + 1) / 2; } void Pushup(int idx) { node[idx].Lnum = node[idx << 1].Lnum; node[idx].rnum = node[idx << 1 | 1].rnum; node[idx].sum = node[idx << 1].sum + node[idx << 1 | 1].sum; //printf("Lnum is %d rnum is %d ",node[idx].Lnum,node[idx].rnum); if(node[idx << 1].rnum <= node[idx << 1 | 1].Lnum) { if(node[idx << 1].Lmx == node[idx << 1].r - node[idx << 1].L + 1) { node[idx].Lmx = node[idx << 1].Lmx + node[idx << 1 | 1].Lmx; } else node[idx].Lmx = node[idx << 1].Lmx; if(node[idx << 1 | 1].rmx == node[idx << 1 | 1].r - node[idx << 1 | 1].L + 1) { node[idx].rmx = node[idx << 1 | 1].rmx + node[idx << 1].rmx; } else node[idx].rmx = node[idx << 1 | 1].rmx; node[idx].sum += cal(node[idx << 1].rmx + node[idx << 1 | 1].Lmx); node[idx].sum -= cal(node[idx << 1].rmx); node[idx].sum -= cal(node[idx << 1 | 1].Lmx); } else { node[idx].Lmx = node[idx << 1].Lmx; node[idx].rmx = node[idx << 1 | 1].rmx; } } void build(int L,int r,int idx) { node[idx].L = L,node[idx].r = r; if(L == r) { node[idx].sum = 1; node[idx].Lmx = node[idx].rmx = 1; node[idx].Lnum = node[idx].rnum = a[L]; return ; } int mid = (L + r) >> 1; build(L,mid,idx << 1); build(mid + 1,r,idx << 1 | 1); Pushup(idx); } void update(int x,int val,int idx) { if(node[idx].L == node[idx].r) { node[idx].Lnum = node[idx].rnum = val; return ; } int mid = (node[idx].L + node[idx].r) >> 1; if(mid >= x) update(x,val,idx << 1); else update(x,val,idx << 1 | 1); Pushup(idx); } LL query(int L,int r,int idx) { if(node[idx].L >= L && node[idx].r <= r) return node[idx].sum; int mid = (node[idx].L + node[idx].r) >> 1; LL ans = 0; if(mid >= L) ans += query(L,r,idx << 1); if(mid < r) ans += query(L,r,idx << 1 | 1); if(mid >= L && mid < r && node[idx << 1].rnum <= node[idx << 1 | 1].Lnum) { int ld,rd; if(node[idx << 1].r - node[idx << 1].rmx + 1 >= L) ld = node[idx << 1].rmx; else ld = node[idx << 1].r - L + 1; if(node[idx << 1 | 1].L + node[idx << 1 | 1].Lmx - 1 <= r) rd = node[idx << 1 | 1].Lmx; else rd = r - node[idx << 1 | 1].L + 1; ans += 1LL * ld * rd; } return ans; } void solve() { int n,q;n = read(),q = read(); for(int i = 1;i <= n;++i) a[i] = read(); build(1,n,1); while(q--) { int id;id = read(); if(id == 1) { int x,y;x = read(),y = read(); update(x,y,1); } else { int L,r;L = read(),r = read(); printf("%lld ",query(L,r,1)); } } } int main() { solve(); //system("pause"); return 0; }
F：首先可以发现，对于每个X，只有0，5，10三种情况。且0比较特殊不用考虑（周围没有'.')
对于5，显然只有1,4可以构造，10也只有1,1,4,4可以构造。
那么可知对于周围有奇数个的'.'一定不存在解。
考虑对于5的情况。那么X周围的两个一定相反，那么连边后就是二分图的染色。
现在考虑10的情况，猜测肯定也存在解。
且我们相邻点继续连边，染色即可。
如果可以染色即为解，不可以那么说明不存在解。
证明不会。
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> pii; typedef tuple<int,int,int> tu; const int N = 2e5 + 5; const int M = 2e4 + 5; const double eps = 1e-10; const LL Mod = 1e9 + 7; #define pi acos(-1) #define INF 1e9 #define dbg(ax) cout << "now this num is " << ax << endl; inline int read() { int f = 1;int x = 0;char c = getchar(); while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } inline long long ADD(long long x,long long y) {return (x + y) % Mod;} inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;} inline long long MUL(long long x,long long y) {return x * y % Mod;} int n,m,deg[505][505],mp[505][505],tot = 0,col[505][505]; pii to[N]; vector<pii> vec; vector<int> G[N]; string a[505]; bool vis[N],f = 0; bool check(int x,int y) { if(x >= 0 && x < n && y >= 0 && y < m && a[x][y] == '.') return true; else return false; } void dfs(int u,int c) { col[to[u].first][to[u].second] = c; vis[u] = 1; //printf("u is %d c is %d x is %d y is %d ",u,c,to[u].first,to[u].second); for(auto v : G[u]) { if(vis[v]) { if(col[to[v].first][to[v].second] == c) f = 1; } else dfs(v,c ^ 1); } } int get_tot(int x,int y) { if(mp[x][y] == 0) { mp[x][y] = ++tot; to[tot].first = x; to[tot].second = y; } return mp[x][y]; } void solve() { n = read(),m = read(); for(int i = 0;i < n;++i) cin >> a[i]; for(int i = 0;i < n;++i) { for(int j = 0;j < m;++j) { if(a[i][j] == 'X') { vec.clear(); if(check(i,j + 1)) deg[i][j]++,vec.push_back(pii{i,j + 1}); if(check(i,j - 1)) deg[i][j]++,vec.push_back(pii{i,j - 1}); if(check(i - 1,j)) deg[i][j]++,vec.push_back(pii{i - 1,j}); if(check(i + 1,j)) deg[i][j]++,vec.push_back(pii{i + 1,j}); if(deg[i][j] == 0) continue; if(deg[i][j] == 1 || deg[i][j] == 3) f = 1; else if(deg[i][j] == 2) { int x = get_tot(vec[0].first,vec[0].second); int y = get_tot(vec[1].first,vec[1].second); G[x].push_back(y); G[y].push_back(x); } else { int x = get_tot(vec[0].first,vec[0].second); int y = get_tot(vec[1].first,vec[1].second); int a = get_tot(vec[2].first,vec[2].second); int b = get_tot(vec[3].first,vec[3].second); G[x].push_back(a);G[a].push_back(x); G[x].push_back(b);G[b].push_back(x); G[y].push_back(a);G[a].push_back(y); G[y].push_back(b);G[b].push_back(y); } } } } for(int i = 1;i <= tot;++i) { if(!vis[i]) { dfs(i,0); } } if(f) printf("NO "); else { printf("YES "); for(int i = 0;i < n;++i) { for(int j = 0;j < m;++j) { if(a[i][j] == 'X') { if(deg[i][j] == 2) { col[i][j] = 5; } else if(deg[i][j] == 4) { col[i][j] = 10; } else col[i][j] = 0; } else if(col[i][j] == 0) col[i][j] = 1; else { col[i][j] = 4; } } } for(int i = 0;i < n;++i) for(int j = 0;j < m;++j) printf("%d%c",col[i][j],j == m - 1 ? ' ' : ' '); } } int main() { solve(); // system("pause"); return 0; }