Codeforces Round #295 Div1 C（Pluses everywhere）

Problem

给一个有 n 位的数 A ，现要求在 A 中插入 k 个加号，使得 A 被分成 k+1 份。比如 1234 ，插入 2 个加号，可变为1+234，12+34，123+4。问所有插入情况得到的数之和（mod 1e9+7），比如1+234=235，12+34=46，123+4=127–>answer=235+46+127

Limits

TimeLimit(ms):3000

MemoryLimit(MB):256

n,k∈[1,105],k<n

Look up Original Problem From here

Solution

用前缀和方法可以求出某个子串所表示的数。

把 A 分成 n 个类，每个类含有若干个长度相同的A的子串，枚举每个类对答案的贡献度，算出即可。实际可枚举子串的长度来求答案。

我用到了前缀和的前缀和 以及 组合数来统计，求组合数用拓展gcd求 Cmn模p 的方法。

Complexity

TimeComplexity:O(n×log2n)

MemoryComplexity:O(n)

My Code

//Hello. I'm Peter.
#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cctype>
#include<ctime>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double ld;
#define peter cout<<"i am peter"<<endl
#define input freopen("data.txt","r",stdin)
#define randin srand((unsigned int)time(NULL))
#define INT (0x3f3f3f3f)*2
#define LL (0x3f3f3f3f3f3f3f3f)*2
#define gsize(a) (int)a.size()
#define len(a) (int)strlen(a)
#define slen(s) (int)s.length()
#define pb(a) push_back(a)
#define clr(a) memset(a,0,sizeof(a))
#define clr_minus1(a) memset(a,-1,sizeof(a))
#define clr_INT(a) memset(a,INT,sizeof(a))
#define clr_true(a) memset(a,true,sizeof(a))
#define clr_false(a) memset(a,false,sizeof(a))
#define clr_queue(q) while(!q.empty()) q.pop()
#define clr_stack(s) while(!s.empty()) s.pop()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define dep(i, a, b) for (int i = a; i > b; i--)
#define repin(i, a, b) for (int i = a; i <= b; i++)
#define depin(i, a, b) for (int i = a; i >= b; i--)
#define pi 3.1415926535898
#define eps 1e-6
#define MOD 1000000007
#define MAXN
#define N 100100
#define M
const ll mod =1e9+7;//这个必须是质数
class Combination_Number{
public://快速求组合数C(n,m)
    ll factor[N];
    void init(){//程序需要先执行这个函数
        factor[0]=1;
        rep(i,1,N){
            factor[i]=(factor[i-1]*i)%mod;
        }
    }
    ll extend_gcd(ll a,ll b,ll &x,ll &y){
        if(b==0){
            x=1;y=0;
            return a;
        }
        ll ans=extend_gcd(b,a%b,x,y);
        ll tmp=x;
        x=y;
        y=tmp-(a/b)*y;
        return ans;
    }
    ll inverse(ll b){
        ll x,y;
        extend_gcd(b,mod,x,y);
        return (x%mod+mod)%mod;
    }
    ll cal(ll n,ll m){//快速求组合数C(n,m)
        ll t1=1,t2=1;
        if(m==0) return 1LL;
        if(m<n/2)m=n-m;
        t1=factor[n];
        t2=factor[n-m]*factor[m]%mod;
        return t1*inverse(t2)%mod;
    }
}C;

int n,k;
char s[N];
ll a[N],b[N],ans=0,tenpower[N],ten;
ll presum[N],sum_presum[N];
ll cal_presum(ll from,ll to,int l){
    ll res=presum[to];
    res-=presum[from-1]*tenpower[l]%mod;
    res=(res%mod+mod)%mod;
    return res;
}
int main(){
    C.init();
    tenpower[0]=1;
    rep(i,1,N){
        tenpower[i]=(tenpower[i-1]*10LL)%mod;
    }
    scanf("%d %d",&n,&k);
    scanf("%s",s);
    rep(i,0,n){
        a[i+1]=s[i]-'0';
    }
    repin(i,1,n){
        presum[i]=(presum[i-1]*10LL+a[i])%mod;
    }
    repin(i,1,n){
        sum_presum[i]=(sum_presum[i-1]+presum[i])%mod;
    }
    if(k==0){
        ans=0;
        repin(i,1,n){
            ans=(ans*10LL+a[i])%mod;
        }
        printf("%lld\n",ans);
        exit(0);
    }
    ans=0;
    repin(l,1,n){
        int r=n-1-l;
        if(r>=k-1){
            ll res1=cal_presum(1,l,l);
            res1=res1*C.cal(r,k-1)%mod;
            ans=(ans+res1)%mod;
            ll res2=cal_presum(n-l+1,n,l);
            res2=res2*C.cal(r,k-1)%mod;
            ans=(ans+res2)%mod;
        }
        r=n-1-(l+1);
        if(k>=1 && r>=k-2){
            int p=2+l-1;
            if(p<=n-1){
                ll res=sum_presum[n-1]-sum_presum[p-1];
                res=(res%mod+mod)%mod;
                res-=sum_presum[n-1-l]*tenpower[l]%mod;
                res=(res%mod+mod)%mod;
                res=res*C.cal(r,k-2)%mod;
                ans=(ans+res)%mod;
            }
        }
    }
    printf("%lld\n",ans);
}