C题题目出错了，unrating，2题就能有很好的名次，只能呵呵了。
/************************************************
* 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;
}
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.\n"; return 0;
}