CodeForces512C-Pluses everywhere-模拟/数学/排列组合模板

经过研究可以发现，每一位的贡献是C(n-2,k-1)+C(n-3,k-1)...C(k-1,k-1)

同时还要注意加号全部在左边的情况。

这里还用了O（n）预处理O(1)组合数的模板。//妙啊。。妙。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long

using namespace std;

const int MOD = 1e9+7;
const int maxn = 3e5+10;

LL f[maxn],inv[maxn],fac[maxn];

void init()
{
    fac[0] = fac[1] = inv[0] = inv[1] = f[0] = f[1] = 1;
    for(int i=2;i<maxn;i++)
    {
        fac[i] = fac[i-1]*i%MOD;
        LL t = MOD/i,k = MOD%i;
        f[i] = (MOD-t)*f[k]%MOD;
        inv[i] = inv[i-1]*f[i]%MOD;
    }
}

LL C(LL n,LL m)
{
    if(n < m || m < 0) return 0;
    return fac[n]*inv[m]%MOD*inv[n-m]%MOD;
}

char s[maxn];
int N,K;
LL save[maxn];

int main()
{
    init();
    while(~scanf("%d%d",&N,&K))
    {
        scanf("%s",s);
        LL base = 1;
        for(int i=1;i<=N;i++)
        {
            save[i] = (save[i-1]+C(N-1-i,K-1)*base%MOD)%MOD;
            base = base*10%MOD;
        }
        LL ans = 0;base = 1;
        for(int i=N-1;i>=0;i--)
        {
            ans += (s[i]-'0')*(save[N-1-i]+C(i,K)*base%MOD)%MOD;
            ans %= MOD;
            base = base*10%MOD;
        }
        printf("%I64d
",ans);
    }
}