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HDOJ 4608 I-number
题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=4608
I-number
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3105 Accepted Submission(s): 1165
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.
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 105.
The following T lines describe all the queries, each with a positive integer x. The length of x will not exceed 105.
Output
Output the I-number of x for each query.
Sample Input
1 202
Sample Output
208
Source
2013 Multi-University Training Contest 1
题意: 给一个正整数x ,要你求最小的y。 使得的y>x 且y的各个位数之和能被10整除。 x的程度不超过 105.
题解: 简单大数加法 .
#include<iostream> #include<string> #define maxn 100000+5 using namespace std; string str; int sum=0,t; int calcbit(string &x) { int s=0; for(int i=0;i<x.size();i++) s+=(x[i]-'0'); return s; } void Add(string &x) { string temp="";char ch; int num[maxn]={0},len=x.size(); for(int i=0;i<len;i++)num[i]=(x[len-i-1]-'0'); num[0]++; for(int i=0;i<len;i++){ if(num[i]>9){ num[i+1]+=num[i]/10; num[i]%=10; } } int i; for(i=len;i>0&&!num[i];)i--; for(;i>=0;i--)temp+=(num[i]+'0'); len=temp.size(); for(int i=0;i<=len/2;i++){ ch=x[i];x[i]=x[len-i-1];x[len-i-1]=ch; } x=temp; } int main() { cin.sync_with_stdio(false); cin>>t; while(t--) { cin>>str; Add(str); sum=calcbit(str); while(sum%10){ Add(str); sum=calcbit(str); } cout<<str<<endl; } return 0; }
HDOJ 4608 I-number
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