ZOJ.3336 Friend Number II【有点水】 2015/10/09

Friend Number II

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Given a positive integer x, let S(x) denotes the sum of all x's digits. Two integersx and y are friend numbers if S(x)=S(y). Here comes the problem: Given a positive integerx, of course it has a lot of friend numbers, find the smallest one which is greater thanx,please.

Input

There are multiple test cases. The first line of input is an integer T (0<T<230) indicating the number of test cases. ThenT test cases follow. Each case is an integer x (0<x<=10¹⁰⁰⁰).

Output

For each test case, output the result integer in a single line.

Sample Input

31219222

Sample Output

2128231

Note: No input data start with digit 0 and you should not output a number starts with 0.

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Author: CAO, Peng
Source: ZOJ Monthly, May 2010

刚开始的时候想错了，以为直接加9就可以了，后来想想不对，仔细想想才明白，从后往前找，找到第一个不为0的减1，再继续往前找，找到第一个不为9的加1(如果找到最前面也没找到一个不为9的，就先输出个1)，然后对此位置之后的串排序输出；

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){    int t,len,i;    char s[1010];    scanf("%d",&t);    while(t--){        scanf("%s",s);        //if( s[0] == '0' ) continue;        len = strlen(s);//printf("%d\n",len);        for( i = len-1 ; i >= 0 ; --i ){            if( s[i] != '0' ){                s[i]--;                break;            }        }        for( --i ; i >= 0 ; --i ){            if( s[i] != '9' ){                s[i]++;                break;            }        }        if( i < 0 ) printf("1");        sort(s+i+1,s+len);        for( i = 0 ; i < len ; ++i )            printf("%c",s[i]);        printf("\n");    }    return 0;}