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Codeforces Round #330 (Div. 2)

2015-11-09 09:44 211 查看
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.\n";

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
, b
;

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.\n";

return 0;
}


　　
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