Codeforces Round #330 (Div. 2)

C题题目出错了，unrating，2题就能有很好的名次，只能呵呵了。

水 A - Vitaly and Night

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/8 星期日 22:41:11
* File Name     :A.cpp
 ************************************************/

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int a[110][220];

int main(void)    {
	int n, m;	cin >> n >> m;
	for (int i=1; i<=n; ++i)	{
		for (int j=1; j<=2*m; ++j)	cin >> a[i][j];
	}
	int ans = 0;
	for (int i=1; i<=n; ++i)	{
        for (int j=1; j<=2*m-1; j+=2)	{
            if (a[i][j] == 1 || a[i][j+1] == 1)	ans++;
        }
	}
	cout << ans;

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}

数学 B - Pasha and Phone

题意：n个数字分成n/k块，每块有k个数字，问n个数字，其中每一块第一个数字不是b[i]且该k个数字组成的数字能整除a[i]的方案数

分析：直接说方法吧，最大的数字比如999999除以a[i]就是所有在99999范围内a[i]的倍数，然后减去以b[i]开头的那些数字就是一块的方案数。我脑子没转过弯，用乘法一个一个乘，当然超时，怎么没想到除呢，还想用DP？ (a[i] < 10 ^ k)，(卒

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/8 星期日 22:41:14
* File Name     :B.cpp
 ************************************************/

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
ll a[N], b[N];

ll get_max(int k)	{
    ll ret = 0;
    for (int i=1; i<=k; ++i)	{
		ret = ret * 10 + 9;
    }
    return ret;
}

ll get_b_max(ll x, int k)	{
	ll ret = x;
	for (int i=1; i<k; ++i)	{
		ret = ret * 10 + 9;
	}
	return ret;
}

ll get_b_min(ll x, int k)	{
	ll ret = x;
	for (int i=1; i<k; ++i)	{
		ret = ret * 10;
	}
	return ret;
}

ll multi_mod(ll a, ll b)	{
    ll ret = 0;
    while (b)	{
		if (b & 1)	{
			ret += a;
			if (ret >= MOD)	ret -= MOD;
		}
		b >>= 1;	a <<= 1;
		if (a >= MOD)	a -= MOD;
    }
    return ret;
}

int main(void)    {
    int n, k;	cin >> n >> k;
	int m = n / k;
	for (int i=1; i<=m; ++i)	{
		cin >> a[i];
	}
	for (int i=1; i<=m; ++i)	{
		cin >> b[i];
	}
	ll ans = 1, mx = get_max (k);
	for (int i=1; i<=m; ++i)	{
        ll cnt = mx / a[i] + 1;
        ll bmn = get_b_min (b[i], k), bmx = get_b_max (b[i], k);
        cnt -= (bmx / a[i] - bmn / a[i]);
        if (bmn % a[i] == 0)    cnt--;
        ans = multi_mod (ans, cnt);
	}
	cout << ans << endl;

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}