Codeforces round #267 (Div. 2)

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D questions Sb was not written out (heavy fog)

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Question C: Can I use wa multiple times? I just need to make a wrong Max transfer! Qaq

 

A. George and accommodation

A and B ask if A is less than or equal to B-2

This...

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>using namespace std;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << #x << " = " << x << endl#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }inline const int max(const int &a, const int &b) { return a>b?a:b; }inline const int min(const int &a, const int &b) { return a<b?a:b; }int main() {int n=getint(), ans=0;while(n--) {int a=getint(), b=getint();if(b-2>=a) ++ans;}print(ans);return 0;}

 

B. Fedor and new game

Q: I will give you m + 1 count so that you can determine whether the binary form of the given number is smaller than or equal to k in number m + 1.

Can you try again ..

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>using namespace std;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << #x << " = " << x << endl#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }inline const int max(const int &a, const int &b) { return a>b?a:b; }inline const int min(const int &a, const int &b) { return a<b?a:b; }const int N=1005;int a[N], my, ans;int main() {int n=getint(), m=getint(), k=getint();for1(i, 1, m) read(a[i]);read(my);for1(i, 1, m) {int tot=0;for3(j, n-1, 0) {if(((1<<j)&a[i])!=((1<<j)&my)) ++tot;}if(tot<=k) ++ans;}print(ans);return 0;}

 

C. George and job

Question: give you n numbers, let you divide K blocks into M contiguous blocks, and then calculate the maximum value of all feasible solutions.

If d [I, j] is set, it indicates the maximum number of I segments divided into J blocks.

D [I, j] = max {d [K, J-1]} + sum [I-m, I], 1 <= k <= I-m

The answer is max {d [I, K]}.

Here is n ^ 3. We want to optimize horizontal n ^ 2.

Set MX [I, j] to max {d [K, J]} 1 <= k <= J

Then transfer

D [I, j] = Mx [I-m, J-1] + sum [I-m, I]

The MX transfer is

MX [I, j] = max {MX [I-1, J], d [I, j]}

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>using namespace std;typedef long long ll;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%I64d", a)#define dbg(x) cout << #x << " = " << x << endl#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }inline const ll getint() { ll r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }inline const ll max(const ll &a, const ll &b) { return a>b?a:b; }inline const ll min(const ll &a, const ll &b) { return a<b?a:b; }const int N=5005;int n, m, k;ll sum[N], d[N], ans, mx[N][N], f[N], a[N];int main() {read(n); read(m); read(k);for1(i, 1, n) read(a[i]), sum[i]=sum[i-1]+a[i];for1(i, m, n) d[i]=sum[i]-sum[i-m];for1(i, m, n) {for3(j, k, 1) {f[j]=mx[i-m][j-1]+d[i];mx[i][j]=max(mx[i-1][j], f[j]);}ans=max(ans, f[k]);}print(ans);return 0;}

 

D. Fedor and essay

String question, it's time for class. Come back to make up at noon ..

 

Codeforces round #267 (Div. 2)