HDU4608:I-number

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I-number Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1524    Accepted Submission(s): 612 Problem Description 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.   Input 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 10 5.   Output Output the I-number of x for each query.   Sample Input 1 202   Sample Output 208   Source 2013 Multi-University Training Contest 1   Recommend liuyiding

=====================================[ 代码 ]=====================================

#include <cstdio>
#include <cstring>
#include <numeric>

using namespace std;

const int kCarryReserve = 1; // 进位预留（只需一位）。

int T;

char buffer[ 100005 ], *x;

int main()
{
    while( scanf( "%d", &T ) == 1 ) while( T-- )
    {
        memset( buffer, '0', kCarryReserve );
        scanf( "%s", x = buffer + kCarryReserve );
        char* last = x + strlen( x );
        
        while( true )
        {
            // 递增。
            char* p = last - 1;
            while( *p == '9' ) { *p-- = '0'; }
            ++*p;
            if( p < x ) { x = p; }
            
            // 求和。
            int sum = accumulate( x, last, '0' * ( x - last ) );
            if( sum % 10 == 0 ) { break; }
        }
        
        printf( "%s\n", x );
    }
    return 0;
}