nowcoder20C 位数差

题意：定义 h(a,b)为在十进制下 a + b 与 a 的位数差，求

题解：观察式子可以发现每个数要往找到b,计算h(a, b)不能每个都算，位数变化很少，所以应该按位数变化分类，计算多少个数位数变化为t可以用数状数组或者线段树维护一下

#include <bits/stdc++.h>
#define maxn 101000
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
ll c[maxn], ans;
int n, a[maxn], b[maxn];
void update(int x,int d){
    for(int i=x;i<maxn;i+=i&(-i))
        c[i] += d;
}
ll sum(int x){
    ll ans = 0;
    for(int i=x;i>=1;i-=i&(-i)) ans += c[i];
    return ans;
}
int bit(ll a){
    if(a == 0) return 1;
    else return (int )log10(a)+1;
}
int main(){
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &a[i]), b[i] = a[i];
    sort(b+1, b+1+n);
    int num = unique(b+1, b+1+n)-(b+1);
    for(int i=1;i<=n;i++) update(lower_bound(b+1, b+1+num, a[i])-b, 1);
    for(int i=1;i<=n;i++){
        update(lower_bound(b+1, b+1+num, a[i])-b, -1);
        for(ll j=10;j<=1e8;j*=10){
            if(j < a[i]) continue;
            int t1 = lower_bound(b+1, b+1+num, j*10-a[i])-b;
            t1--;
            int t2 = lower_bound(b+1, b+1+num, j-a[i])-b;
            t2--;
            ans += (sum(t1)-sum(t2))*(bit(j)-bit(a[i]));
        }

    }
    printf("%lld
", ans);
    return 0;
}