Codeforces Round #524 Div. 2 翻車記
阿新 • • 發佈：2018-11-25
round 太差 lib esp 條件 有一個 out print ace 
A：簽到。room裏有一個用for寫的，hack了一發1e8 1，結果用了大概600+ms跑過去了。慘絕人寰。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘View Code)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;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; } int n,m; int main() { /*#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif*/ n=read(),m=read(); cout<<(n*2-1)/m+1+(n*5-1)/m+1+(n*8-1)/m+1; return 0; }
B：討論一發即可。
#include<iostream> #includeView Code<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 1010 char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;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; } int n; int main() { /*#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif*/ n=read(); while (n--) { int x=read(),y=read(); int p=y;if (y-x+1&1) y--; int ans=x&1?(y-x+1>>1):-(y-x+1>>1); if (p-x+1&1) ans+=p&1?-p:p; printf("%d\n",ans); } return 0; }
C：對矩形求個交，冷靜一下瞎算算就行了。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 1010 char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;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; } int T,n,m; ll area(int x1,int y1,int x2,int y2){if (x1>x2||y1>y2) return 0;return 1ll*(x2-x1+1)*(y2-y1+1);} ll calcwhite(int x1,int y1,int x2,int y2) { ll s=area(x1,y1,x2,y2); if (x1+y1&1) return s>>1; else return s+1>>1; } ll calcblack(int x1,int y1,int x2,int y2){return area(x1,y1,x2,y2)-calcwhite(x1,y1,x2,y2);} int main() { /*#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif*/ T=read(); while (T--) { n=read(),m=read(); int x1=read(),y1=read(),x2=read(),y2=read(); int x3=read(),y3=read(),x4=read(),y4=read(); int x5=max(x1,x3),y5=max(y1,y3),x6=min(x2,x4),y6=min(y2,y4); ll white=calcwhite(1,1,n,m),black=calcblack(1,1,n,m); ll c1=calcblack(x1,y1,x2,y2);white+=c1,black-=c1; black+=calcwhite(x3,y3,x4,y4),white-=calcwhite(x3,y3,x4,y4); if (area(x5,y5,x6,y6)) black+=calcblack(x5,y5,x6,y6),white-=calcblack(x5,y5,x6,y6); printf("%I64d %I64d\n",white,black); } return 0; }View Code
D：考慮答案為i時的最小和最大分裂次數是多少，各種遞推式求通項到最後能搗鼓出來兩個式子。顯然答案若存在與n相差不會很大，暴力枚舉check即可。當然式子不能直接算，需要各種亂搞防溢出。數學太差推一年。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll unsigned long long #define N 1010 char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} ll read() { ll x=0,f=1;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; } int T,n,m; int main() { /*#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif*/ T=read(); while (T--) { ll n=read(),m=read(); int ans=-1; for (int i=(n>=63?n-63:0);i<n;i++) { ll k=1; for (int j=1;j<=n-i+1;j++) k<<=1; k-=n-i;k-=2; if (k<=m) { if (n+i+1>=63) {ans=i;break;} ll k=1;for (int j=1;j<=n-i-1;j++) k<<=1;k--;ll t=k+1; for (int j=1;j<=n+i+1;j++) { k<<=1; if (k>=3*m+2) {ans=i;break;} } if (ans!=-1) break; t<<=1;if (t>=3*m+2) {ans=i;break;} t<<=1;if (t>=3*m+2) {ans=i;break;} k+=t;if (k>=3*m+2) {ans=i;break;} t=1; for (int j=1;j<=2*i;j++) { t<<=1; if (t>=3*m+2) {ans=i;break;} } if (ans!=-1) break; k+=t;if (k>=3*m+2) {ans=i;break;} } } if (ans==-1) printf("NO\n"); else printf("YES %d\n",ans); } return 0; }View Code
E：這個條件比較顯然的等價於子矩形中每一行都能重排成回文串（即出現次數為奇數的字母不超過一個）且對稱行的字母組成相同。預處理每一行的每一段是否合法以及字母出現次數的哈希值，暴力枚舉子矩形左右端點，對上下端點的每段合法區間馬拉車即可。完全沒時間寫了。
F：沒看
小號打的。result：rank 139 rating +92
Codeforces Round #524 Div. 2 翻車記