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Codeforces Round #267 (Div. 2) A B C

史值拥 2023-03-03 阅读 90


题目链接:​​http://codeforces.com/contest/467​​​
A 水
代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <queue>
#include <set>

using namespace std;

int n;
int p, q;

int main()
{
while (cin >> n)
{
int ans = 0;
for (int i = 1; i <= n; i++)
{
cin >> p >> q;
if (q - p >= 2) ans++;
}
cout << ans << endl;
}
return 0;
}

B 暴力一发
代码:

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

using namespace std;

int x[10005];

int test(int x)
{
int ans = 0;
while (x)
{
if (x & 1) ans++;
x = x >> 1;
}
return ans;
}

int main()
{
int n, m, k;
while (~scanf("%d%d%d", &n, &m, &k))
{
m++;
for (int i = 1; i <= m; i++)
scanf("%d",&x[i]);
int ans = 0;
for (int i = 1; i < m; i++)
{
if (test(x[i] ^ x[m]) <= k) ans++;
}
printf("%d\n", ans);
}
return 0;
}

C dp ;
解法:dp[i][j]表示 (前i个 ,分为j组的时候最优解。
代码:

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

using namespace std;

typedef long long LOL;

int n, m, k;
LOL p[5010], dp[5010][5010];

int main()
{
while (cin >> n >> m >> k)
{
memset(p,0,sizeof(p));
memset(dp,0,sizeof(dp));
for (int i = 1; i <= n; i++)
{
cin >> p[i];
p[i] += p[i-1];
}
//dp[i][j]表示 (前i个 ,分为j组的时候最优解
for (int i = m; i <= n; i++)
for (int j = 1; j <= k; j++)
{
dp[i][j] = max(dp[i-1][j],dp[i-m][j-1] + p[i]-p[i-m]);
}
cout << dp[n][k] << endl;
}
return 0;
}


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