25 lines
630 B
C
25 lines
630 B
C
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#include <stdio.h>
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int main(void){
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unsigned long long n, c = 0, x, pow = 1, lx = 0;
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scanf("%llu",&n);
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while(n){ //当n不等于0的时候才继续循环
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x = n % 10; //将n现在的最低位赋值给x
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n /= 10;
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if(n){//当n不等于0的时候,就是判断现在的处理的位是否到了最后一位
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if(x == 1){
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c += (lx+1) + n * pow; //总数加上现在的位上出现1的次数
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}
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else c += (x>0?n+1:n) * pow; //同上
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}
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else if(x < 2)
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c += lx+1;
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else c += pow;
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lx += x * pow;//记录x位后面的位，作为以后x=1的时候的计算依据
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pow *= 10;
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}
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printf("%llu\n",c);
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return 0;
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}
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