【组合数学】 Codeforces Round #295 (Div. 1) C - Pluses everywhere

考虑每一位数作为各位，十位，百位。。对答案的贡献。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 100005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

char s[maxn];
LL f[maxn];
LL g[maxn];
int n, m;

void read()
{
	scanf("%d%d", &n, &m);
	scanf("%s", s+1);
	f[0] = 1;
	for(int i = 1; i <= n; i++) f[i] = f[i-1] * i % mod;
	g[n] = powmod(f[n], mod - 2);
	for(int i = n-1; i >= 0; i--) g[i] = g[i+1] * (i+1) % mod;
}

inline LL C(int k1, int k2)
{
	if(k2 > k1) return 0;
	return f[k1] * g[k2] % mod * g[k1 - k2] % mod;
}

void work()
{
	LL ans = 0, sum = 0, mul = 1;
	for(int i = n; i >= 1; i--) {
		LL t = s[i] - '0';
		ans = (ans + (sum + mul * C(i-1, m) % mod) * t) % mod;
		sum = (sum + C(i-2, m-1) * mul) % mod;
		mul = mul * 10 % mod;
	}
	printf("%I64d\n", ans);
}

int main()
{
	read();
	work();
	
	return 0;
}