[AGC001A] BBQ Easy

为了让最小的值不影响更大的值，只搭配次小的。

#include <bits/stdc++.h>
#define d(x) cerr << #x << " => " << x << endl
#define D cerr << "Passed [" << __FUNCTION__ << "] on line " << __LINE__ << endl
#define dd(x) cerr << #x << endl
#define ms(x) memset(x, 0, sizeof x)
using namespace std;
int rd() {
    int f = 1, val = 0; char ch = getchar();
    while(!isdigit(ch)) f = (ch == '-' ? -1 : 1), ch = getchar();
    while(isdigit(ch)) val = val * 10 + ch - '0', ch = getchar();
    return f * val;
}
const int N = 1e5 + 5;
int a[N];
signed main() {
    int n = rd(); 
    for(int i = 1; i <= 2 * n; i++) a[i] = rd();

    sort(a + 1, a + 2 * n + 1);
    int ans = 0;
    for(int i = 1; i <= 2 * n; i += 2) ans += a[i];
    cout << ans << endl;
    return 0;
}

[AGC001B] Mysterious Light

递归，除前两步操作外，本质是遍历一个循环的平行四边形。

#include <bits/stdc++.h>
#define d(x) cerr << #x << " => " << x << endl
#define D cerr << "Passed [" << __FUNCTION__ << "] on line " << __LINE__ << endl
#define dd(x) cerr << #x << endl
#define ms(x) memset(x, 0, sizeof x)
#define int long long
using namespace std;
int rd() {
    int f = 1, val = 0; char ch = getchar();
    while(!isdigit(ch)) f = (ch == '-' ? -1 : 1), ch = getchar();
    while(isdigit(ch)) val = val * 10 + ch - '0', ch = getchar();
    return f * val;
}

int solve(int a, int b) {
    if(b == 0) return 0;
    if(a < b) return solve(b, a);
    if(a % b == 0) return 2 * (a - a % b) - b;
    return 2 * (a - a % b) + solve(b, a % b);
}

signed main() {
    int n = rd(), x = rd();
    cout << n + solve(x, n - x) << endl;
    return 0;
}

[AGC002A] 过水

[AGC002C] Knot Puzzle

如果有相邻的两段绳子比 L 长，我们可以从绳子的两端依次剪掉。

#include <bits/stdc++.h>
#define d(x) cerr << #x << " => " << x << endl
#define D cerr << "Passed [" << __FUNCTION__ << "] on line " << __LINE__ << endl
#define dd(x) cerr << #x << endl
#define ms(x) memset(x, 0, sizeof x)
using namespace std;
int rd() {
    int f = 1, val = 0; char ch = getchar();
    while(!isdigit(ch)) f = (ch == '-' ? -1 : 1), ch = getchar();
    while(isdigit(ch)) val = val * 10 + ch - '0', ch = getchar();
    return f * val;
}

const int N = 1e5 + 5;
int a[N];
void solve() {
    int n = rd(), L = rd();
    for(int i = 1; i <= n; i++) {
        a[i] = rd();
    }

    int flag = true;
    for(int i = 2; i <= n; i++) {
        if(a[i - 1] + a[i] >= L) {
            puts("Possible");
            for(int j = 1; j < i - 1; j++) {
                printf("%d\n", j);
            }

            for(int j = n - 1; j >= i; j--) {
                printf("%d\n", j);
            }

            printf("%d\n", i - 1);
            flag = false;
            break;
        }
    }

    if(flag) {
        puts("Impossible");
    }

}

signed main() {
    int T = 1; 
    while(T--) solve();
    return 0;
}