1018 Subnumbers(35 分)

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前缀和统计 预处理
枚举每个非零位的贡献扫一次就好了

#include <bits/stdc++.h>
using namespace std;
#define MOD (int)(1e9+7)
#define ll long long
#define db(a) (cout<<"->"<<a<<endl)
ll ppow(ll a,ll n){
    ll ret=1;
    while(n){
        if(n&1) ret=ret*a%MOD;
        a*=a%MOD;
        n>>=1;
    }
    return ret;
}
ll now[102000];
ll pre[102000];
ll fac10[102000];
int main(){
    string s;
    cin>>s;
    ll len=s.size();
    fac10[1]=1;
    for(ll i=2;i<=101000;i++){
        fac10[i]=(fac10[i-1]*10+1)%MOD;
    }
    s='#'+s;
    for(ll i=1;i<=len;++i){
        now[i]=s[i]-'0';
    }
    for(ll i=1;i<=len;++i){
        if(now[i]) pre[i]=pre[i-1]+1;
        else pre[i]=pre[i-1];
    }
    ll ans=0;
    for(ll i=1;i<=len;++i){
        if(now[i]) ans=(ans+now[i]*fac10[len-i+1]%MOD*pre[i]%MOD)%MOD;
    }
    cout<<ans<<endl;
    return 0;
}

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转载自blog.csdn.net/qq_38677814/article/details/82491822