2020-05-27 — 习题训练三题解

https://vjudge.net/contest/375686#overview

A - Sorted Adjacent Differences

从小到大排序，最后一个减最前面一个的差最大，并且倒数第二个减去第一个比刚刚的差值小，循环着输出就行了

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
int t, n, a[N], b[N];
int main()
{
    cin >> t;
    while(t--){
        cin >> n;
        for(int i = 1;i <= n; ++ i) cin >> a[i];
        sort(a + 1, a + 1 + n);
        int s = 1, e = n, k = n;
        while(s <= e)
        {
            b[k--] = a[e--];
            b[k--] = a[s++];
        }
        for(int i = 1;i <= n; ++ i) cout << b[i] << " ";
        cout << endl;
    }

}

 B - Powered Addition

记录最大与最小的差值，位运算数位数即可

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

template <typename T>

inline void read(T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) return ;
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    sgn = (c == '-') ? -1:1;
    ret = (c == '-') ? 0:(c - '0');
    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return ;
}
template <class T>
inline void out(T x) {
    if (x > 9) out(x / 10);
    putchar(x % 10 + '0');
}
int t;
void solve(){
        
    int n, res = 0;
    read(n);
    vector<ll> a(n);
    
    for(int i = 0;i < n; ++ i) read(a[i]);
    
    ll l= a[0], m = 0;
    
    for(int i = 0;i < n; ++ i){
        l = max(a[i], l);
        m = max(l - a[i], m);
    }
    while(m) m >>= 1, res ++;
    
    out(res);
    puts("");
}
int main()
{
    ios_base::sync_with_stdio(false);
    read(t);
    while(t--) solve();
}

C - Middle Class

 从大到小排序， 依次取平均，有一个小于目标就失败了

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

void solve()
{
    int n, x;
    cin >> n >> x;
    
    vector<int> v(n);
    for(int i = 0;i < n; ++ i) cin >> v[i];
    sort(v.begin(),v.end(), greater<int>());
    
    ll sum = 0;
    for(int i = 0;i < n; ++ i){
        sum += v[i];
        if(sum / (i + 1) < x){
            cout << i << "\n";
            return ;
        }
    }
    cout << n << "\n";
}
int main()
{
    ios_base::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--) solve();
}

D - Circle of Monsters

 记录被b[i[伤害后的血量，求个和(sum)。需要有一个对着满血的怪开枪，sum = (求和)a[i] - b[i-1] (a[i] 前面那个), 所以应该加上最小的b[i-1](a[i] 前面那个), 也就是min{a[i] - c[i}], c[i] = a[i] - b[i-1](a[i] 前面那个);

 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

template <typename T>

inline void read(T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) return ;
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    sgn = (c == '-') ? -1:1;
    ret = (c == '-') ? 0:(c - '0');
    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return ;
}
void solve(){
    int n;
    read(n);
    vector<ll> a(n), b(n), c(n);
    for(int i = 0;i < n; ++ i){
        read(a[i]), read(b[i]);
    }
    
    c[0] = max(0ll, a[0] - b[n-1]);
    ll sum = c[0], k = a[0] - c[0];
    
    for(int i = 1;i < n; ++ i){
        c[i] = max(0ll, a[i] - b[i-1]);
        sum += c[i];
        k = min(k, a[i] - c[i]);
    }
    
    printf("%lld\n",sum+k);
}
int main()
{
    ios_base::sync_with_stdio(false);
    int t;
    read(t);
    while(t--) solve();
}

E - Kind Anton

a[i]前面有1可以解决b[i] > a[i], 有-1可以解决b[i] < a[i]; 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

template <typename T>

inline void read(T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) return ;
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    sgn = (c == '-') ? -1:1;
    ret = (c == '-') ? 0:(c - '0');
    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return ;
}
template <class T>
inline void out(T x) {
    if (x > 9) out(x / 10);
    putchar(x % 10 + '0');
}
void solve(){
        
    int n;
    read(n);    
    vector<int> a(n), b(n);
    for(int i = 0;i < n; ++ i) read(a[i]);
    for(int i = 0;i < n; ++ i) read(b[i]);
    
    int cnt1 = 0, neg_cnt1 = 0;
    if(a[0] != b[0]){
        puts("NO");
        return ;
    }
    if(a[0] == 1) cnt1 ++;
    if(a[0] == -1) neg_cnt1 ++;
    
    for(int i = 1;i < n; ++ i){
        if(cnt1 && neg_cnt1){
            puts("YES");
            return ;
        }
        if(b[i] > a[i] && cnt1 == 0){
            puts("NO");
            return ;
        } 
        if(b[i] < a[i] && neg_cnt1 == 0){
            puts("NO");
            return ;
        }
        
        if(a[i] == 1) cnt1 ++;
        if(a[i] == -1) neg_cnt1 ++;
        
    }
    puts("YES");
}
int main()
{
    ios_base::sync_with_stdio(false);
    int t;
    read(t);
    while(t--) solve();
}