Problem Description
11 Dimensions is a cute contestant being talented in math. One day, 11 Dimensions came across a problem but didn't manage to solve it. Today you are taking training here, so 11 Dimensions turns to you for help.
You are given a decimal integer S with n bits s1s2…sn(0≤si≤9), but some bits can not be recognized now(replaced by "?''). The only thing you know is that Sis a multiple of a given integer m.
There may be many possible values of the original S, please write a program to find the k-th smallest value among them. Note that you need to answer q queries efficiently.
Input
The first line of the input contains an integer T(1≤T≤10000), denoting the number of test cases.
In each test case, there are three integers n,m,q(1≤n≤50000,2≤m≤20,1≤q≤100000) in the first line, denoting the length of S, the parameter m, and the number of queries.
In the second line, there is a string s of length n, denoting the given decimal integer S. It is guaranteed that si is either an integer within [0,9] or ``?'', and s1 is always an integer within [1,9].
For the next q lines, each line contains an integer ki(1≤ki≤1018), denoting each query.
It is guaranteed that ∑n≤500000 and ∑q≤10^6.
Output
For each query, print a single line containing an integer, denoting the value of S. If the answer exists, print Smod(10^9+7) instead, otherwise print ``-1''.
Sample Input
1
5 5 5
2??3?
1
2
3
100
10000
Sample Output
20030
20035
20130
24935
-1
多次查询第K小的使所给数字%m==0的数字是多少。
先进行dp,dp[i][j]表示填了i个之后取模等于J的情况有多少种。因为可知k<=1e18的,所以::
然后开始从大到小尝试放数就可以了。
#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = 50005;
const int mod = 1e9 + 7;
#define ll long long
#define itn int
const ll INF = 1e18 + 5;
int pow_mod(ll a, ll b, ll c) {
ll ans = 1;
while (b > 0) {
if (b & 1) ans = ans * a % c;
a = a * a % c;
b /= 2;
}
return ans;
}
ll dp[maxn][20];
int pos[maxn];
ll pm[maxn], pmod[maxn];
string res;
int n, m, q, cnt = 0;
void init() {
for (int i = 1; i <= cnt; i++) {
pm[i] = pow_mod(10, pos[i], m);
pmod[i] = pow_mod(10, pos[i], mod);
}
for (int i = 0; i <= cnt; i++) {
for (int j = 0; j < 20; j++)
dp[i][j] = 0;
}
dp[0][0] = 1;
}
int main() {
int te;
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> te;
while (te--) {
cin >> n >> m >> q;
cin >> res; int len = n; cnt = 0;
ll tmpans = 0, summ = 0;//不算问号的数字的取余mod,不算问号的数字的取余m
for (int i = 0; i < len; i++) {
tmpans *= 10; summ *= 10;
if (res[i] != '?') {
tmpans += res[i] - '0';
summ += res[i] - '0';
}
summ %= m; tmpans %= mod;
}
for (int i = len - 1; i >= 0; i--) {
if (res[i] == '?') {
pos[++cnt] = len - i - 1;
}
}
summ = (m - summ) % m;
init();
for (int i = 1; i <= cnt; i++) {
for (int j = 0; j <= 9; j++) {
int now = j * pm[i] % m;
for (int k = 0; k < m; k++) {
int ne = (now + k) % m;
dp[i][ne] += dp[i - 1][k];
dp[i][ne] = min(dp[i][ne], INF);
}
}
}
while (q--) {
ll ans = tmpans;
ll k; cin >> k;
if (dp[cnt][summ] < k) {
cout << "-1\n";
continue;
}
int now = summ;
for (int i = min(cnt, 30); i >= 1; i--) {
for (int j = 0; j <= 9; j++) {
int ne = ((now - j * pm[i] % m) + m) % m;
if (dp[i - 1][ne] < k) {
k -= dp[i - 1][ne];
}
else {
ans += j * pmod[i] % mod;
ans %= mod;
now = ne;
break;
}
}
}
cout << ans % mod << "\n";
}
}
}