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+;
 const int maxn = 3e5+;

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

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

 LL C(LL n,LL m)
 {
     if(n < m || m < ) return ;
     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 = ;
         for(int i=;i<=N;i++)
         {
             save[i] = (save[i-]+C(N--i,K-)*base%MOD)%MOD;
             base = base*%MOD;
         }
         LL ans = ;base = ;
         for(int i=N-;i>=;i--)
         {
             ans += (s[i]-'')*(save[N--i]+C(i,K)*base%MOD)%MOD;
             ans %= MOD;
             base = base*%MOD;
         }
         printf("%I64d\n",ans);
     }
 }