C. Vasya And The Mushrooms

題解：拿筆畫他的行走路線，你會發現可以前綴和預處理一些東西出來。

const int N = 300005;

int n;
ll a[N], b[N], dp[2][N], sp[2][N], sum[N];

int get_a() {

    dp[0][0] = 0;
    for (int i = 1; i < n; ++i) {
        dp[0][i] = dp[0][i - 1] + 1ll * i * a[i];
    }

    int t = n;
    dp[1][n - 1] = dp[0][n - 1] + 1ll * t * b[n - 1
 
];

    for (int i = n - 2; ~i; --i) {
        t++;
        dp[1][i] = dp[1][i + 1] + 1ll * t * b[i];
    }
}

int get_b() {

    sp[1][0] = b[1];
    for (int i = 1; i < n; ++i) sp[1][i] = sp[1][i - 1] + 1ll * (i + 1) * b[i];

    int t = n + 1;
    sp[0][n - 1] = sp[1][n - 1] + 1ll * t * a[n - 1];

    
 
for (int i = n - 2; ~i; --i) {
        t++;
        sp[0][i] = sp[0][i + 1] + 1ll * t * a[i];
    }
}

void Inite() {
    for (int i = n - 1; ~i; --i) sum[i] = sum[i + 1] + a[i] + b[i];
}

void solve() {
    ll ans = max(dp[1][0], sp[0][1]);
    int last = 0;
    ll tp = 0;
    for (int i = 0; i < n; ++i) {
        
 
if(!last) {
            tp += 1ll * 2 * i * a[i] + 1ll * (2 * i + 1) * b[i];
            if(i + 1 != n) ans = max(ans, tp + 1ll * i * sum[i + 1] + sp[0][i + 1] - sp[1][i]);
        }
        else {
            tp += 1ll * 2 * i * b[i] + 1ll * (2 * i + 1) * a[i];
            if(i + 1 != n) ans = max(ans, tp + 1ll * (i + 1) * sum[i + 1] + dp[1][i + 1] - dp[0][i]);
        }
        last ^= 1;
    }
    printf("%I64d\n", max(ans, tp));
}

D. Vasya And The Matrix

題解：可以這麽構造，除了第一行，第一列填上以外，其它的行和列都填0，所以主要是算map[1][1]改填幾，或者填最後一行和最後一列。

 ans[n][m] = b[m] ^ (x ^ a[n]);

Educational Codeforces Round 48 (Rated for Div. 2) - 賽後補題