Codeforces Round #675 (Div. 2)

题目传送门

A. Fence

签到

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
ll a, b, c;
 
 
inline void solve(int T)
{
    cin >> a >> b >> c;
    cout << a + b + c - 1 << endl;
}
 
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
 
}

B. Nice Matrix

模拟

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
ll n, m;
ll ans;
ll a[110][110];
 
inline void solve(int T)
{
 
    cin >> n >> m;
    rep(i, 1, n) rep(j, 1, m) cin >> a[i][j];
    ans = 0;
    rep(i, 1, n / 2) rep(j, 1, m / 2) {
        ll tmp = abs(a[i][j] - a[n - i + 1][j]) + abs(a[i][j] - a[i][m - j + 1]) + abs(a[i][j] - a[n - i + 1][m - j + 1]);
        tmp = min(tmp, abs(a[i][j] - a[n - i + 1][j]) + abs(a[n - i + 1][j] - a[i][m - j + 1]) + abs(a[n - i + 1][j] - a[n - i + 1][m - j + 1]));
        tmp = min(tmp, abs(a[i][m - j + 1] - a[n - i + 1][j]) + abs(a[i][j] - a[i][m - j + 1]) + abs(a[i][m - j + 1] - a[n - i + 1][m - j + 1]));
        tmp = min(tmp, abs(a[n - i + 1][m - j + 1] - a[n - i + 1][j]) + abs(a[n - i + 1][m - j + 1] - a[i][m - j + 1]) + abs(a[i][j] - a[n - i + 1][m - j + 1]));
        ans += tmp;
    }
    if(n & 1) rep(j, 1, m / 2) ans += abs(a[n / 2 + 1][j] - a[n / 2 + 1][m - j + 1]);
    if(m & 1) rep(i, 1, n / 2) ans += abs(a[i][m / 2 + 1] - a[n - i + 1][m / 2 + 1]);
    cout << ans << endl;
}
 
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
 
}

C. Bargain

找规律，假设有长度为6的数: $a_1,a_2,a_3,a_4,a_5,a_6$

然后找每一位的系数，

$a_6: frac{(5 + 1) * 5}{2} * 1$

$a_5: frac{(4 + 1) * 4}{2} * 10 + 1 * 1$

$a_4: frac{(3 + 1) * 3}{2} * 100 + 2 * 10 + 1 * 1$

$a_3: frac{(2 + 1) * 2}{2} * 1000 + 3 * 100 + 2 * 10 + 1 * 1$

$a_2: frac{(2 + 1) * 2}{2} * 10000 + 4 * 1000 + 3 * 100 + 2 * 10 + 1 * 1$

$a_1: ...$

规律很容易看出了

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
ll n;
char s[200010];
ll cnt, tmp, ans, mod = 1e9 + 7;
 
inline void solve(int T)
{
    cin >> s + 1;
    n = strlen(s + 1);
    ans = tmp = 0;
    cnt = 1;
    for(int i = n; i; i--) {
        ans = (ans + 1ll * (s[i] - '0') * i * (i - 1) / 2 % mod * cnt % mod) % mod;
        ans = (ans + 1ll * (s[i] - '0') * tmp) % mod;
        tmp = (tmp + 1ll * (n - i + 1) * cnt % mod) % mod;
        cnt = cnt * 10 % mod;
        
    }
    cout << ans << endl;
}
 
 
int main()
{
    // ios_base::sync_with_stdio(0);
    // cin.tie(0);
    // cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    // cin >> T;
    rep(i, 1, T) solve(i);
 
}