1049. Counting Ones (30)计算个位数出现次数.

传送门:https://www.patest.cn/contests/pat-a-practise/1049#include <iostream>
#include <vector>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <string.h>
#include <unordered_map>
#include <map>
using namespace std;
int a[10]={0};
void deal(int n,int value){
if(n<=0)
return;
int one,ten;
one = n%10;
n/=10;
ten=n;
for (int i=0; i<=one; ++i) {//个位上出现的数统计下来
a[i]+=value;
}
while (ten) {
a[ten%10]+= (one+1)*value;
ten/=10;
}
for (int i=0; i<10; ++i) {
a[i]+=value*n;
}
a[0]-=value;//第一位是0的情况排除.
deal(n-1,value*10);
}
int main(){
int n;
scanf("%d",&n);
deal(n,1);
printf("%d\n",a[1]);
return 0;
}