Educational Codeforces Round 48

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C.Vasya And The Mushrooms

思路很簡單，走法有一個統一形式就是先上下走，然后到某個位置左右一個來回。然后就推一下，后邊那段的遞推式子，枚舉改變走法的位置即可。看出做法之后發現要推個式子，于是跑去寫D了。。。然后D一開始思路錯了。。。涼啊

#include <bits/stdc++.h> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) #define pb push_back #define PII pair<int,int> #define MP make_pair #define fr first #define sc second #define mem(WM) memset(WM,0,sizeof(WM)) typedef long long ll; typedef unsigned long long ull; const int N = 3e5 + 7; using namespace std; int n; ll a[N][2],b[N][2],sum[N][2],s[N],A[N][2],num[N][2]; int main() {scanf("%d",&n);rep(i,1,n)scanf("%I64d",&a[i][0]);rep(i,1,n)scanf("%I64d",&a[i][1]);rep(i,1,n) s[i] = s[i-1] + a[i][0]+a[i][1];sum[n][0] = a[n][1];sum[n][1] = a[n][0];per(i,n-1,1) {sum[i][0] = (s[n]-s[i]) + sum[i+1][0] + ((n-i+1)*2-1)*a[i][1];sum[i][1] = (s[n]-s[i]) + sum[i+1][1] + ((n-i+1)*2-1)*a[i][0];}ll ans = sum[1][0];int sy=0,cc=0;rep(i,1,n) {A[i][sy] = A[i-1][sy] + cc*a[i][sy];num[i][sy] = cc;sy^=1;++cc;ans = max(ans,A[i][sy]);A[i][sy] = A[i][sy^1] + cc*a[i][sy];num[i][sy] = cc;ans = max(ans,A[i][sy]);++cc;}sy = 1;rep(i,1,n) {ll tmp = A[i][sy] + sum[i+1][sy] + (s[n] - s[i])*(num[i][sy]+1);ans = max(ans,tmp);sy^=1;}printf("%I64d\n",ans);return 0; }

D.Vasya And The Matrix

給定橫縱的異或值，讓你求一個符合條件的矩陣。拆位之后，通過一行一列控制異或值，再檢查以下最后一列/行是否滿足條件即可。

#include <bits/stdc++.h> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) #define pb push_back #define PII pair<int,int> #define MP make_pair #define fr first #define sc second #define mem(WM) memset(WM,0,sizeof(WM)) typedef long long ll; typedef unsigned long long ull; const int N = 1e5 + 7; const ll mod = 1e9 + 7; using namespace std;int n, m; ll mp[111][111],a[111],b[111],r[111],c[111]; ll ans[111][111]; int solve() {rep(i,1,m-1) {if(c[i]==0) rep(j,1,n) mp[j][i]=c[i];else {mp[1][i] = 1;rep(j,2,n) mp[j][i]=0;}}rep(i,1,n) {ll x=0;rep(j,1,m-1) x^=mp[i][j];if(x==r[i]) mp[i][m] = 0;else mp[i][m] = 1;}ll x=0;rep(i,1,n)x^=mp[i][m];if(x==c[m]) return 1;rep(i,1,n-1) {if(!r[i]) rep(j,1,m) mp[i][j]=r[i];else {mp[i][1] = 1;rep(j,2,m) mp[i][j] = 0;}}rep(i,1,m){ll x=0;rep(j,1,n-1)x^=mp[j][i];if(x==c[i]) mp[n][i] = 0;else mp[n][i] = 1;}x=0;rep(i,1,m)x^=mp[n][i];if(x==r[n]) return 1;return -1; } int tmp[111]; int main() {scanf("%d%d",&n,&m);rep(i,1,n) scanf("%I64d",&a[i]);rep(i,1,m) scanf("%I64d",&b[i]);rep(x,0,30) {rep(i,1,n) r[i] = !!(a[i]&(1<<x));rep(i,1,m) c[i] = !!(b[i]&(1<<x));int t = solve();if(t == -1)return puts("NO"),0;rep(i,1,n)rep(j,1,m)ans[i][j] += (mp[i][j]<<x);}puts("YES");rep(i,1,n) {rep(j,1,m) printf("%I64d ",ans[i][j]);puts("");}return 0; }

E. Rest In The Shades

大概做法懂了，對每個詢問的點它和下方的那個光源的軌跡會形成一個三角形，二分求x軸上兩條邊中間的覆蓋區域即可。這個可以預處理前綴和。二分兩個邊界，兩邊要特判一下相交。然后相似三角形即可求出在軌跡上的投影。。。

轉載于:https://www.cnblogs.com/RRRR-wys/p/9417149.html

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