AtCoder Beginner Contest 247 C D E题解
https://atcoder.jp/contests/abc247/tasks/abc247_c
递归即可解决：
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<string> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> typedef long long ll; using namespace std; typedef unsigned long long int ull; #define maxn 200005 #define ms(x) memset(x,0,sizeof(x)) #define Inf 0x7fffffff #define inf 0x3f3f3f3f const int mod = 1e9 + 7; #define pi acos(-1.0) #define pii pair<int,int> #define eps 1e-5 #define pll pair<ll,ll> #define lson 2*x #define rson 2*x+1 long long qupower(int a, int b) { long long ans = 1; while (b) { if (b & 1)ans = ans * a; b >>= 1; a = a * a; } return ans; } inline int read() { int an = 0, x = 1; char c = getchar(); while (c > '9' || c < '0') { if (c == '-') { x = -1; } c = getchar(); } while (c >= '0'&&c <= '9') { an = an * 10 + c - '0'; c = getchar(); } return an * x; } const int N = 65540; int a[N]; void sol(int x){ if(x==1){printf("1 ");return;} sol(x-1); printf("%d ",x); sol(x-1); } int main(){ //ios::sync_with_stdio(false); int n; n = read(); sol(n); }
https://atcoder.jp/contests/abc247/tasks/abc247_d
不能用queue来模拟，复杂度过不去。我们只需要记录每一段的位置（前缀和维护），以及该段的value。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<string> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> typedef long long ll; using namespace std; typedef unsigned long long int ull; #define maxn 200005 #define ms(x) memset(x,0,sizeof(x)) #define Inf 0x7fffffff #define inf 0x3f3f3f3f const int mod = 1e9 + 7; #define pi acos(-1.0) #define pii pair<int,int> #define eps 1e-5 #define pll pair<ll,ll> #define lson 2*x #define rson 2*x+1 long long qupower(int a, int b) { long long ans = 1; while (b) { if (b & 1)ans = ans * a; b >>= 1; a = a * a; } return ans; } inline int read() { int an = 0, x = 1; char c = getchar(); while (c > '9' || c < '0') { if (c == '-') { x = -1; } c = getchar(); } while (c >= '0'&&c <= '9') { an = an * 10 + c - '0'; c = getchar(); } return an * x; } int Q; //queue<ll> q; ll ans = 0; const int N = 2e5+5; ll v[N],pos[N]; ll cnt=1; ll cur_p = 1; ll real_p = 0; int main(){ //ios::sync_with_stdio(false); Q = read(); while(Q--){ int q; q = read(); if(q==1){ ll x,c; scanf("%lld %lld", &x,&c); pos[cnt] = pos[cnt-1]+c; v[cnt] = x; cnt+=1; } else{ ll c; scanf("%lld", &c); ll next_p = real_p+c; //cout<<"next pos:"<<next_p<<endl; ans=0; for(;cur_p<cnt;cur_p++){ if(pos[cur_p]<=next_p){ ans+= ll(v[cur_p]*(pos[cur_p]-real_p)); real_p = pos[cur_p]; } else{ ans+= ll(v[cur_p]*(next_p-real_p)); real_p = next_p; break; } } //cout<<real_p<<endl; printf("%lld\n", ans); } } }
https://atcoder.jp/contests/abc247/tasks/abc247_e
给定一个序列，以及两个数 X,Y. 问有多少子序列（连续的）满足：该序列的最小值为Y，最大值为X。
注意到的一点是：如果有一个数字不满足条件，即 a>X or a<Y，此时包含该数字的序列都不满足，所以我们可以根据不满足条件的数字来分割成小串。现在问题回到了如何对于满足条件的序列来统计数量。
我们使用双指针 L,R. 开始的时候移动右指针 R。当找到最小值和最大值时，即可break来统计数量。具体来说，如果区间 [ L, W ]满足条件，那么区间 [ L, i ]，W< i < R都满足条件。最后移动左指针。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<string> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> typedef long long ll; using namespace std; typedef unsigned long long int ull; #define maxn 200005 #define ms(x) memset(x,0,sizeof(x)) #define Inf 0x7fffffff #define inf 0x3f3f3f3f const int mod = 1e9 + 7; #define pi acos(-1.0) #define pii pair<int,int> #define eps 1e-5 #define pll pair<ll,ll> #define lson 2*x #define rson 2*x+1 long long qupower(int a, int b) { long long ans = 1; while (b) { if (b & 1)ans = ans * a; b >>= 1; a = a * a; } return ans; } inline int read() { int an = 0, x = 1; char c = getchar(); while (c > '9' || c < '0') { if (c == '-') { x = -1; } c = getchar(); } while (c >= '0'&&c <= '9') { an = an * 10 + c - '0'; c = getchar(); } return an * x; } const int N = 2e5 + 5; int a[N]; int X,Y; int n; ll res=0; ll cal(vector<int> &vec){ ll ans=0; int l = 0, r = vec.size()-1; int lp=l,rp=l; int cnt_max=0, cnt_min = 0; //int len = r-l+1; while(lp<=r){ while(rp<=r && (cnt_max==0 || cnt_min==0)){ if(vec[rp]==X)cnt_max+=1; if(vec[rp]==Y)cnt_min+=1; rp+=1; } if(cnt_max>0 && cnt_min>0){ ans = ans+ll(r-rp+2); } if(vec[lp]==X)cnt_max-=1; if(vec[lp]==Y)cnt_min-=1; lp+=1; } return ans; } void sol(){ int i=0; int st=0,ed=0; while(i<n){ // split to subproblem vector<int> vec; while(i<n && a[i]<=X && a[i]>=Y){vec.push_back(a[i]);i+=1;} if(vec.size()>0){ res+= cal(vec); } else i+=1; } } int main(){ //ios::sync_with_stdio(false); n = read(); X = read(); Y = read(); for(int i=0;i<n;i++) a[i] = read(); sol(); printf("%lld\n", res); }
