Codeforces Round #675 (Div. 2)

題目鏈接：http://codeforces.com/contest/1422

A題

簽到題

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MODE = 1e9 + 7;
string s;
int main () {
	cin >> s;
	ll n = s.size();
	ll temp = 0; 
	for(int i = 0; i < n; ++i) {
		temp += s[i] - '0';
	}	
	ll po = 1;
	ll sum = 0;
	for(ll i = n - 1; i >= 0; --i) {
		ll x =s[i] - '0';
		temp -= x;
		sum += temp * (n - i) % MODE * po % MODE;  //之后 
		sum += x * i * (i + 1) / 2 % MODE * po % MODE; //算上這位 
		sum %= MODE;
		po *= 10;
		po %= MODE;
	}
	cout << sum << endl;
} 

 B題

思維題

對稱四元組全部變成四個中第二大的數

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll get(ll a, ll b,ll c, ll d) {
	vector<ll> v(4);
	v[0] = a;
	v[1] = b;
	v[2] = c;
	v[3] = d;
	sort(v.begin(), v.end());
	return v[1];
} 
int main () {
	int t;
	cin >> t;
	while(t--) {
		ll v[110][110];
		int n, m;
		cin >> n >> m;
		ll sum = 0;
		for(int i = 0; i < n; ++i) {
			for(int j = 0; j < m; ++j) {
				cin >> v[i][j];
			}
		}
		for(int i = 0; i < n; ++i) {
			for(int j = 0; j < m; ++j) {
				sum += abs(v[i][j] - get(v[i][j], v[n - 1 - i][j], v[i][m - 1 - j], v[n - 1 -i][m - 1 - j]));
			}
		} 
		cout << sum << endl;
	}
} 

 C題

數論題

有點兒靈性得看出每一位對於答案的貢獻

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MODE = 1e9 + 7;
string s;
int main () {
	cin >> s;
	ll n = s.size();
	ll temp = 0; 
	for(int i = 0; i < n; ++i) {
		temp += s[i] - '0';
	}	
	ll po = 1;
	ll sum = 0;
	for(ll i = n - 1; i >= 0; --i) {
		ll x =s[i] - '0';
		temp -= x;
		sum += temp * (n - i) % MODE * po % MODE;  //temp代表的是x之前所有數的和，乘以po之后代表，對於其的貢獻 
		sum += x * i * (i + 1) / 2 % MODE * po % MODE; //i(i+1)/2是求和。代表的是x選擇后左邊不同位數可以選擇的情況總數 
		sum %= MODE;
		po *= 10;
		po %= MODE;
	}
	cout << sum << endl;
}