A - Xenny and Alternating Tasks
题解
枚举第一天是谁做，将两个答案取(min)即可。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
int t, n, a[20003], b[20003];
int main()
{
//File("XENTASK");
t = gi();
while (t--)
{
n = gi();
for (int i = 1; i <= n; i+=1) a[i] = gi();
for (itn i = 1; i <= n; i+=1) b[i] = gi();
itn ans = 0, sum = 0;
for (itn j = 1; j <= n; j+=1)
{
if (j & 1) sum = sum + a[j];
else sum = sum + b[j];
}
for (int j = 1; j <= n; j+=1)
{
if (j & 1) ans = ans + b[j];
else ans = ans + a[j];
}
printf("%d
", min(ans, sum));
}
return 0;
}
B - Bear and Extra Number
题解
将数列排序，遍历数组元素，然后分类讨论：
- 如果是(a_i = a_{i+1}) ，那么输出(a_i)。
- 如果(a_{i+1}-a_i>1) ：
- 因为题目说有唯一解，因此(i)只能为(n-1)或(1)。
- 若(i)为(n-1)，则输出(a_n)。
- 若(i)为(1)，则输出(a_1)。
- 因为题目说有唯一解，因此(i)只能为(n-1)或(1)。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
int t, n, a[100003], ans, sum;
int main()
{
//File("EXTRAN");
t = gi();
while (t--)
{
n = gi();
for (itn i = 1; i <= n; i+=1) a[i] = gi();
sort(a + 1, a + 1 + n);
for (int i = 1; i < n; i+=1)
{
if (a[i + 1] == a[i]) {printf("%d
", a[i]); break;}
if (a[i + 1] - a[i] > 1)
{
if (i == n - 1) {printf("%d
", a[n]); break;}
else if (i == 1) {printf("%d
", a[1]); break;}
}
}
}
return 0;
}
C - Cooking Schedule
题解
二分。
注意(check)怎么写。
第一次做时没想到二分，很遗憾。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define int long long
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
int n, m, t, k, ans, kkk, c[1000003];
char s[1000003];
int CCC(itn x)
{
int u = 0;
for (int i = 1; i <= n; i+=1)
{
if (i % 2 == x)
{
if (s[i] == '0') ++u;
}
else {if (s[i] == '1') ++u;}
}
return u;
}
bool check(int x)
{
if (x != 1)
{
int u = 0;
for (itn i = 1; i <= kkk; i+=1) if (c[i] > x) u = u + c[i] / (x + 1);
return u <= k;
}
else {int u = min(CCC(1), CCC(0)); return u <= k;}
}
signed main()
{
//File("SCHEDULE");
t = gi();
while (t--)
{
int l = 1, r = 0, uuu = 0; kkk = 0;
n = gi(), k = gi();
scanf("%s", s + 1);
for (int i = 1; i <= n; i+=1)
{
if (s[i] != s[i + 1])
{
c[++kkk] = i - uuu; uuu = i; r = max(r, c[kkk]);
}
}
while (l < r)
{
int mid = (l + r) >> 1;
if (check(mid)) r = mid;
else l = mid + 1;
}
printf("%lld
", r);
}
return 0;
}
D - Subtree Removal
题解
简单树形(DP)。
看似很难，实际代码很短。
计算出每棵子树的大小，与(-X)取(max)即可。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <vector>
#define int long long
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
int t, n, x, fa[100003], ans, sum, k, b[100003], sz[100003];
vector <int> e[100003];
int dfs(int u, int fa)
{
int o = b[u];
for (int i = e[u].size() - 1; ~i; i-=1)
{
int v = e[u][i];
if (v == fa) continue;
o = o + dfs(v, u);
}
return max(o, -x);
}
signed main()
{
//File("SUBREM");
t = gi();
while (t--)
{
n = gi(), x = gi();
int sum = 0;
for (itn i = 1; i <= n; i+=1) b[i] = gi(), sz[i] = b[i], sum = sum + b[i], e[i].clear();
for (itn i = 1; i < n; i+=1) {int u = gi(), v = gi(); e[u].push_back(v), e[v].push_back(u);}
printf("%lld
", dfs(1, -1));
}
return 0;
}
E - Dish Owner
题解
并查集简单题。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <vector>
#include <queue>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
itn t, n, s[10003], fl, x, y, Q;
itn a[10003], fa[10003];
int getf(itn u) {if (u == fa[u]) return u; return fa[u] = getf(fa[u]);}
inline void unionn(int u, int v) {
if (a[u] > a[v])
{
fa[v] = u;
}
else if (a[v] > a[u])
{
fa[u] = v;
}
}
int main()
{
//File("DISHOWN");
t = gi();
while (t--)
{
n = gi();
for (itn i = 1; i <= n; i+=1) s[i] = gi(), a[i] = s[i], fa[i] = i;
Q = gi();
for (itn i = 1; i <= Q; i+=1)
{
fl = gi();
if (fl)
{
x = gi();
printf("%d
", getf(x));
}
else
{
x = gi(), y = gi();
int X = getf(x), Y = getf(y);
if (X == Y) puts("Invalid query!");
else unionn(X, Y);
}
}
}
return 0;
}
F - Triplets
题解
数学题。
大暴力很好写，考虑如何优化。
预处理出(a)、(c)序列的前缀和。
逐一枚举(b)序列上的数，找出(a)、(c)序列中比当前枚举到的数小的数。
计算出其对答案的贡献即可。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define int long long
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
const int mod = 1000000007;
itn t, n, p, q, r, a[100003], b[100003], c[100003], x, y, z;
itn s1 = 1, s2 = 1, s3 = 1;
signed main()
{
//File("SUMQ");
t = gi();
while (t--)
{
p = gi(), q = gi(), r = gi();
for (int i = 1; i <= p; i+=1) a[i] = gi();
for (int i = 1; i <= q; i+=1) b[i] = gi();
for (itn i = 1; i <= r; i+=1) c[i] = gi();
sort(a + 1, a + 1 + p); sort(b + 1, b + 1 + q); sort(c + 1, c + 1 + r);
int p1 = 1, p2 = 1, c1 = 0, c2 = 0, s1 = 0, s2 = 0, ans = 0;
for (int i = 1; i <= q; i+=1)
{
while (p1 <= p && a[p1] <= b[i])
{
s1 = (s1 + a[p1]) % mod;
++c1, ++p1;
}
while (p2 <= r && c[p2] <= b[i])
{
s2 = (s2 + c[p2]) % mod;
++c2, ++p2;
}
ans = (ans % mod + c1 * c2 % mod * b[i] % mod * b[i] % mod) % mod;
ans = (ans % mod + b[i] * (s1 * c2 % mod + s2 * c1 % mod) % mod) % mod;
ans = (ans % mod + s1 * s2 % mod) % mod;
}
printf("%lld
", ans % mod);
}
return 0;
}
总结
这次练习第一次做的时候做得一般般。
做题的策略要更好一点。
细节地方要注意。