CF1422C Bargain (DP+Thinking+Mathematics)

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Title:

Given an integer $n(1\le len(n)<10^5$ , where$len\left(n\right)$ means$The\; number\; of\; digits\; of\; n$ )) , now it is stipulated that each round of Vova can take out a continuous number of digits, and the remaining digits are the score of the round. The result can have leading$0$ . In particular, if Vova takes all numbers, the round scores$0$ .

Now ask you to find the sum of the scores of all the different ways of taking the pair $10^9+$The value after modulo$7.\; In\; particular,\; if\; there\; are\; many\; different\; ways\; to\; get\; the\; same\; number,\; the\; score\; will\; be\; counted\; multiple\; times.$

analyze:

First, let's analyze the number in the i-th place to see how much he can contribute:
There are two (three) ways that he can contribute:

• delete the number in front of the i bit
• delete the number after the i bit
• Delete a continuous segment containing i-bits (delete does not contribute)

Then we analyze the contribution made by deleting the number in front of the i-bit: First of all, it is certain that the contribution of the i-bit is unchanged before deleting the i-bit, which is $str\left[i\right]\ast 10^{len-i+1}$ Then we only need to calculate how many times, just look at how many kinds of intervals there are in the previous paragraph? Select 2 numbers from the previous (i-1) numbers as the left and right intervals, which is (i-1)*i/2.

Then look at the contribution made by the number after deleting the i-bit: the same as the analysis of the number before deleting the i-bit, the difference is that the contribution of the i-bit changes, we find that his contribution changes like this 1, 20, 300, 4000..., because each When there is one more behind him, there is one more choice.

/// 欲戴皇冠，必承其重。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;
typedef unsigned long long ull;

#define x first
#define y second
#define PI acos(-1)
#define inf 0x3f3f3f3f
#define lowbit(x) ((-x)&x)
#define debug(x) cout << #x << ": " << x << endl;

const int MOD = 998244353;
const int mod = 1e9+7;
const int N = 5e5 + 10;
const int dx[] = {
    
    0, 1, -1, 0, 0, 0, 0};
const int dy[] = {
    
    0, 0, 0, 1, -1, 0, 0};
const int dz[] = {
    
    0, 0, 0, 0, 0, 1, -1};
int day[] = {
    
    0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

ll n, m,p;
string str;

void solve()
{
    
    
    cin>>str;
    ll len=str.size();
    str=" "+str;
    ll ans=0;
    ll sum=0;
    p=1;
    for(ll i=len;i>=1;i--){
    
    
        ll now = (i-1)*i/2;
        ans = (ans+p*(str[i]-'0')%mod*now%mod)%mod;///只删前面】
        ans = (ans + sum * (str[i]-'0') % mod ) % mod;///只删后面
        sum = (sum + (len-i+1) *p %mod) % mod;
        p = p * 10 % mod;
    }
    printf("%lld\n",ans);
}    
int main()
{
    
    
    ll t = 1;
    ///scanf("%lld", &t);
    while(t--)
    {
    
    
        solve();
    }
    return 0;
}