A. Equator(模拟)

找权值的中位数,直接模拟。。

代码写的好丑qwq。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAXN = 1e6 + ;
inline int read() {
char c = getchar(); int x = , f = ;
while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
while(c >= '' && c <= '') x = x * + c - '', c = getchar();
return x * f;
}
int N;
int a[MAXN];
main() {
#ifdef WIN32
// freopen("a.in", "r", stdin);
#endif
int N = read(), sum = ;
for(int i = ; i <= N; i++) a[i] = read(), sum += a[i];
int now = ;
for(int i = ; i <= N; i++) {
now += a[i];
if(!(sum & )) {
if(now >= sum / ) {
printf("%d", i); return ;
}
} else {
if(now > sum / ) {
printf("%d", i); return ;
}
}
}
}

A

B. Students in Railway Carriage(贪心)

直接贪心放,如果是偶数的话肯定是两种各放一半

如果是奇数的话,应该在都放一半的基础上,把多的放在最后一个

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAXN = 1e6 + ;
inline int read() {
char c = getchar(); int x = , f = ;
while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
while(c >= '' && c <= '') x = x * + c - '', c = getchar();
return x * f;
}
int N;
char s[MAXN];
main() {
#ifdef WIN32
// freopen("a.in", "r", stdin);
#endif
int N = read(), A = read(), B = read();
if(A < B) swap(A, B);
scanf("%s", s + );
int last = ;
int ans = ;
s[N + ] = '*';
for(int i = ; i <= N + ; i++) {
if(s[i] == '*') {
int len = i - last - ;
ans += min(len / , A) + min(len / , B);
A -= min(len / , A); B -= min(len / , B);
if((len & ) && A > ) A--, ans++;
if(A < B) swap(A, B);
last = i;
}
} printf("%d", ans);
}

B

C. Make a Square(数论,暴力)

首先把所有平方数预处理出来

然后对于输出的数的各个位分解出来,

枚举所有的平方数,算出最少需要删几个

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define int long long
using namespace std;
const int MAXN = 1e6 + ;
inline int read() {
char c = getchar(); int x = , f = ;
while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
while(c >= '' && c <= '') x = x * + c - '', c = getchar();
return x * f;
}
int po[MAXN], tot = , N, P[MAXN], Pnum = ;
int check(int val) {
int times = ;
for(int i = ; i <= Pnum; i++) {
if(P[i] == val % ) {
val /= , times++;
}
if(val == ) {
return Pnum - times;
}
}
if(val != ) return -;
else return Pnum - times;
}
main() {
#ifdef WIN32
// freopen("a.in", "r", stdin);
#endif
for(int i = ; ; i++) {
po[++tot] = i * i;
if(po[tot] > * 1e9) break;
}
N = read();
int x = N;
while(x) P[++Pnum] = x % , x /= ;
int ans = 1e9;
for(int i = ; i <= tot; i++) {
int num = check(po[i]);
if(num != -) ans = min(ans, num);
}
printf("%I64d", ans == 1e9 ? - : ans);
}

C

D. Merge Equals(堆)

用优先队列维护每个数的位置和权值(权值相同的时候位置小的在前面)

然后每次取出前两个,如果相同就合成完再放回去,否则统计答案

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define int long long
using namespace std;
const int MAXN = 1e6 + ;
inline int read() {
char c = getchar(); int x = , f = ;
while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
while(c >= '' && c <= '') x = x * + c - '', c = getchar();
return x * f;
}
int N;
struct Node {
int pos, val;
bool operator < (const Node &rhs) const {
return val == rhs.val ? pos > rhs.pos : val > rhs.val;
}
};
priority_queue<Node>q;
int ans[MAXN];
void work() {
while(q.size() >= ) {
Node x = q.top(); q.pop();
Node y = q.top(); q.pop();
if(x.val != y.val) {q.push(y); ans[x.pos] = x.val; continue;}
q.push((Node){y.pos, x.val * });
}
while(q.size() != ) ans[q.top().pos] = q.top().val, q.pop();
}
main() {
#ifdef WIN32
// freopen("a.in", "r", stdin);
#endif
N = read();
for(int i = ; i <= N; i++)
q.push((Node){i, read()});
work();
int num = ;
for(int i = ; i <= N; i++)
if(ans[i] != )
num++;
printf("%I64d\n", num);
for(int i = ; i <= N; i++)
if(ans[i] != )
printf("%I64d ", ans[i]);
}

D

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