CodeForces Round 521 div3
代码：
#include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL mod = (int)1e9+7; const int N = 1e5 + 100; int main(){ int T, a, b, c; scanf("%d", &T); while(T--){ scanf("%d%d%d", &a, &b, &c); LL t = a - b; t = t * (c/2); if(c&1) t += a; printf("%I64d\n", t); } return 0; }
代码：
#include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL mod = (int)1e9+7; const int N = 1e5 + 100; int a[N]; int main(){ int n; scanf("%d", &n); a[0] = 0; a[n+1] = 0; for(int i = 1; i <= n; ++i) scanf("%d", &a[i]); int ans = 0; for(int i = 1; i <= n; ++i){ if(a[i-1] == 1 && a[i+1] == 1 && a[i] == 0){ a[i+1] = 0; ans++; } } printf("%d\n", ans); return 0; }
题解：模拟， 判断的时候注意 如果是用数组可能会下标越界。
代码：
#include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL mod = (int)1e9+7; const int N = 1e6 + 100; int a[N]; int vis[N]; vector<int> vc; int main(){ int n; scanf("%d", &n); LL sum = 0; for(int i = 1; i <= n; ++i){ scanf("%d", &a[i]); ++vis[a[i]]; sum += a[i]; } for(int i = 1; i <= n; ++i){ sum -= a[i]; --vis[a[i]]; if(sum%2 == 0 && sum/2 < N){ int t = sum/2; //cout << i <<" "<< t << endl; if(vis[t]) vc.pb(i); } sum += a[i]; ++vis[a[i]]; } printf("%d\n", vc.size()); for(auto i : vc){ printf("%d ", i); } return 0; }
题解：二分次数，然后输出。
代码：
#include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL mod = (int)1e9+7; const int N = 1e6 + 100; int a[N]; int b[N]; int n, k, m = 0; bool check(int x){ int ret = 0; for(int i = 1; i <= m; ++i){ ret += a[b[i]]/x; } return ret >= k; } int main(){ scanf("%d%d", &n, &k); for(int i = 1, t; i <= n; ++i){ scanf("%d", &t); ++a[t]; } for(int i = 1; i < N; ++i){ if(a[i]){ b[++m] = i; } } int l = 1, r = n; while(l <= r){ int mid = l+r >> 1; if(check(mid)) l = mid+1; else r = mid-1; } l--; for(int i = 1, c = 1; c <= k; ){ if(a[b[i]] >= l){ a[b[i]] -= l; c++; printf("%d ", b[i]); } else i++; } return 0; }
题解：将每种类型的话题存在一起， 然后sort一下，把小的排前面，然后跑一下背包就好了。
代码：
#include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL mod = (int)1e9+7; const int N = 2e5 + 100; int a[N]; int b[N]; int dp[N]; int main(){ int n; scanf("%d", &n); for(int i = 1; i <= n; ++i) scanf("%d", &a[i]); sort(a+1, a+1+n); int m = 0; for(int i = 1; i <= n; ++i){ if(a[i] == a[i-1]) b[m]++; else b[++m] = 1; } sort(b+1, b+1+m); memset(dp, -inf, sizeof(dp)); dp[0] = 0; for(int i = 1; i <= m; ++i){ for(int j = b[i]; j > 0; --j){ dp[j] = max(dp[j], j); if(j%2 == 0) dp[j] = max(dp[j], dp[j/2] + j); } } int ans = 0; for(int i = 1; i < N; ++i) ans = max(ans, dp[i]); printf("%d\n", ans); return 0; }
题解：dp[i][u] 表示 处理到i 之后选了u个点， 他的最大价值是多少。
画画图之后就发现发现， dp[i][u] 可以从 dp[i - x][u-1] 到 dp[i-1][u-1] 转移过来。
所以对于 dp[i][u] 来说我们需要找到 dp[i-x][u-1] 到 dp[i-1][u-1] 里面的最大值。
我一开始是想用线段树搞，写好了之后MLE了......
后来也发现线段树有太多浪费的点了，然后用set， 可能操作太多了， 然后TLE了。。。。。
然后把set改成优先队列就过了，但是跑的太慢了。。。。
最后改成了单调栈， 这个东西不带log 就跑到200ms内了，我一开始是想用这个东西，然后忘了怎么写，就搞了这么多奇奇怪怪的东西。
代码：
#include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<LL,int> pll; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL mod = (int)1e9+7; const int N = 5100; int a[N]; int n, k, x; deque<pll> dq[N]; int main(){ scanf("%d%d%d", &n, &k, &x); for(int i = 1; i <= n; ++i) scanf("%d", &a[i]); dq[0].push_back({0,0}); LL ans = -1; for(int i = 1; i <= n; ++i){ for(int j = x-1; j >= 0; --j){ while(!dq[j].empty() && dq[j].front().se < i-k) dq[j].pop_front(); if(dq[j].empty()) continue; LL val = dq[j].front().fi; val += a[i]; while(!dq[j+1].empty() && dq[j+1].back().fi <= val) dq[j+1].pop_back(); dq[j+1].push_back({val,i}); if(j+1 == x && i+k > n) ans = max(ans, val); } } printf("%lld\n", ans); return 0; }
