Codeforces #105 DIV2 ABCDE
2024-08-31 18:19:59
开始按照顺序刷刷以前的CF。
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == ? b : gcd(b, a % b);}
int a[],d;
int main()
{
while (cin>>a[]>>a[]>>a[]>>a[]>>d)
{
int ans = ;
for (int i = ; i <= d; i++)
{
for (int j = ; j < ; j++)
if (i % a[j] == )
{
ans ++;
break;
}
}
cout<<ans<<endl;
}
return ;
}
B贡献无数发WA。注意相遇的条件 直接double处理就可以。一直以为按照相应直接直接跳跃相应距离 直接int处理
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == ? b : gcd(b, a % b);}
int vp,vd,t,f,c;
int pos[];
int main()
{
while (cin>>vp>>vd>>t>>f>>c)
{
if (vp >= vd) {puts("");continue;}
int thev = vd - vp;
double pospri = vp * t;int ans = ;
while (pospri < c)
{
double ti = pospri / (double)thev;
if ((double) vd * ti >= c) break;
ans ++;
pospri = pospri + f * vp +(double) * ti * vp;
// printf("%d\n",pospri);
if (pospri >= c) break;
}
cout <<ans <<endl;
}
return ;
}
C 坑点比较多。注意b等于0的情况
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == ? b : gcd(b, a % b);}
int n,a,b;
int num[];
int main()
{
while (cin >> n >> a >> b)
{
if (a > n || b > n){ puts("-1");continue;}
if (a > && b == )
{
if (n - a < ) {puts("-1");continue;}
else
{
printf("1 1 ");
for (int i = ,k = ,j = ; j <= n && i <= a ; k++,i++) printf("%d ",k);
for (int i = a + ; i <= n; i++) printf("%d ",a + );
puts("");
}
continue;
}
for (int i = ; i < ; i++) num[i] = ;
int cas = ;
int sum = ;
for (int i = ; i <= b + ; i++)
{
num[i] = sum + cas;
sum += num[i];
}
cas = ;
for (int i = b + ; i <= a + b + ;i ++)
{
num[i] = num[i - ] + cas;
}
for (int i = ; i <= n; i++)
if (i == ) cout << num[i];
else cout << ' ' << num[i];
puts("");
}
return ;
}
D DP
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == ? b : gcd(b, a % b);}
int w,b;
double dp[][];
void init()
{
for (int i = ; i < ; i++)
{
dp[i][] = 1.0;
dp[][i] = 0.0;
}
for (int w = ; w < ; w++)
for (int b = ; b < ; b++)
{
dp[w][b] += 1.0 * w / (w + b);
if (b >= )
dp[w][b] += dp[w][b - ] * 1.0 * b / (w + b) * 1.0 * (b - ) / (w + b - ) * 1.0 * (b - ) / (w + b - );
if (b >= )
dp[w][b] += dp[w - ][b - ] * 1.0 * b / (w + b) * 1.0 * (b - ) / (w + b - ) * w / (w + b - );
}
}
int main()
{
init();
while (scanf("%d%d",&w,&b) != EOF)
printf("%.9lf\n",dp[w][b]);
return ;
}
E DP预处理+DP
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == ? b : gcd(b, a % b);}
#define MAXN 110
#define MAXD 10010
int sum[MAXN][MAXD],dp[MAXN][MAXD],res[MAXN][MAXD];
int N,M;
int num[MAXN];
int main()
{
while (scanf("%d%d",&N,&M) != EOF)
{
memset(dp,,sizeof(dp));
memset(res,,sizeof(res));
for (int i = ; i <= N; i++)
{
scanf("%d",&num[i]);
sum[i][] = ;
for (int j = ; j <= num[i];j++)
{
int tmp;
scanf("%d",&tmp);
sum[i][j] = sum[i][j - ] + tmp;
}
res[i][] = ;
for (int j = ; j <= num[i]; j++)
for (int k = ; k <= j; k++)
res[i][j] = max(res[i][j],sum[i][k] + sum[i][num[i]] - sum[i][num[i] - (j - k)]);
}
for (int i = ; i <= N; i++)
for (int j = ; j <= M; j++)
for (int k = ; k <= num[i] && k <= j; k++)
dp[i][j] = max(dp[i][j],max(dp[i - ][j],dp[i - ][j - k] + res[i][k]));
printf("%d\n",dp[N][M]);
}
return ;
}
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