HDU 4608 I-number（模拟）

I-number

Time Limit: 5000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

The I-number of x is defined to be an integer y, which satisfied the the conditions below:
1.  y>x;
2.  the sum of each digit of y(under base 10) is the multiple of 10;
3.  among all integers that satisfy the two conditions above, y shouble be the minimum.
Given x, you\'re required to calculate the I-number of x.

输入

An integer T(T≤100) will exist in the first line of input, indicating the number of test cases.
The following T lines describe all the queries, each with a positive integer x. The length of x will not exceed 10⁵.

输出

Output the I-number of x for each query.

示例输入

1202

示例输出

208

提示

来源

2013 Multi-University Training Contest 1

 

我不得不说这道题真的很坑，我要是不搜题解估计是毁掉了 本来题意很简单，就是给定一个x（由于可能很大用字符串模拟），求出比x大的且各位数字之和为10的倍数的数，

要求输出符合上诉条件的最小数。but,这个x竟然是可以有前导0的（它没说）也就是说输入 00202 输出 00208

 

代码略挫QAQ

#include <stdio.h>#include <iostream>#include <algorithm>#include <string.h>#include <cstdio>#include <cctype>using namespace std;char x[100010];int digitsum(){int sum=0;for(int i=0;i<strlen(x);i++)sum+=(x[i]-'0');return sum;}int is_o(){for(int i=0;i<strlen(x);i++)if(x[i]!='0') return 0;return 1;}int main(){ int i,T; cin>>T; getchar(); while(T--) { cin>>x; int s=-99; while(s%10){int p=strlen(x)-1;int len=strlen(x);x[p]++;while(x[p]>'9'){x[p--]-=10;if(p>=0)x[p]++;}if(is_o()){x[0]='1';for(i=1;i<=len;i++)x[i]='0';x[i]='\0';}s=digitsum();}cout<<x<<endl; }  return 0;}