/* - - - - - - - - - - - - - - -
    User :      VanishD
    problem :   [bzoj1361]
    Points :    
- - - - - - - - - - - - - - - */
# include <bits/stdc++.h>
# define    ll      long long
# define    inf     0x3f3f3f3f
# define    M       200010
# define    N       1010
using namespace std;
int read(){
    int tmp = 0, fh = 1; char ch = getchar();
    while (ch < '0' || ch > '9'){ if (ch == '-') fh = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9'){ tmp = tmp * 10 + ch - '0'; ch = getchar(); }
    return tmp * fh;
}
struct INT{
    int len, num[N];
    void print(){
        for (int i = len; i > 0; i--)
            printf("%d", num[i]);
        printf("\n");
    }
}tnp, f[N], nxt;
INT operator -(INT &a, INT &b){
    nxt = a; int i;
    for (i = 1; i <= nxt.len; i++) nxt.num[i] -= b.num[i];
    for (i = 1; i < nxt.len; i++)
        if (nxt.num[i] < 0)
            nxt.num[i] += 10, nxt.num[i + 1] -= 1;
    while (nxt.num[nxt.len] == 0)
        nxt.len--;
    return nxt;
}
INT operator *(INT &a, int b){
    nxt = a; int i;
    for (i = 1; i <= nxt.len; i++) nxt.num[i] *= b;
    for (i = 1; i <= nxt.len || nxt.num[i] != 0; i++)
        nxt.num[i + 1] += nxt.num[i] / 10, nxt.num[i] %= 10;
    nxt.len = i - 1;
    return nxt;
}
INT operator /(INT &a, int b){
    memset(nxt.num, 0, sizeof(nxt.num));
    nxt.len = 0; int i, tmp = 0;
    for (i = a.len; i >= 1; i--){
        tmp = tmp * 10 + a.num[i];
        nxt.num[i] = tmp / b;
        nxt.len = max(nxt.len, (nxt.num[i] > 0) ? i : 0);
        tmp %= b;
    }
    return nxt;
}
char mp[M];
int n, m, k;
int main(){
//  freopen("bzoj1361.in", "r", stdin);
//  freopen("bzoj1361.out", "w", stdout);
    n = read(), m = read(), k = read();
    scanf("\n%s", mp + 1);
    tnp.len = 1, tnp.num[1] = 1;
    for (int i = 1; i <= k; i++){
        tnp = tnp * 2;
        f[i] = tnp;
        for (int j = 1; j < i; j++)
            if (i % j == 0) f[i] = f[i] - f[j];
    }
    f[k] = f[k] / k; 
    f[k].print();
    for (int i = 1; i <= m; i++) mp[i] -= '0';
    for (int i = m + 1; i <= n; i++)
        mp[i] = mp[(i - 1) % m + 1];
    mp[n]++;
    int tmp = n;
    while (mp[tmp] == 2){
        mp[tmp--] = 0;
        mp[tmp]++;
    }
    for (int i = 1; i <= tmp; i++)
        printf("%d", mp[i]);
    printf("\n");
    return 0;
}

图片：“老爹炸了！！！”

图片爆了。。。。

第一问容斥原理+高精可以做

第二问贪心做，尽量多地复制原串，再加1，最后去掉结尾多余的0.

占地盘~图片呢？

没图

- -||

哇！！！

好神奇啊！！！

图片看不到耶...#

好多楼啊好多楼

已经这么多楼了...仰望

我要看片。。。。。。占楼

图片去哪了

wa...

我说吖……图片都没有怎么做啊……

额，先来占下，这道题还100%阿，不好意思交了