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): 1369 Accepted Submission(s): 554
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
1202
Sample Output
208
Source
2013 Multi-University Training Contest 1
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解题思路:看给的时间,应该知道可以直接暴力解决,直接连续自加直到所有数字之和为10的倍数为止。本题涉及到达数,不能直接用某个数据类型存储,用数组模拟存贮即可,看到有大神用数组模拟万进制存储,在下不才,还处于小鸟阶段,只能做到数组十进制存储,容易理解。注意,本题坑爹的地方在:如果输入得数据中有前导零时,输出也要注意前导零,还要注意进位。
解题思路:看给的时间,应该知道可以直接暴力解决,直接连续自加直到所有数字之和为10的倍数为止。本题涉及到达数,不能直接用某个数据类型存储,用数组模拟存贮即可,看到有大神用数组模拟万进制存储,在下不才,还处于小鸟阶段,只能做到数组十进制存储,容易理解。注意,本题坑爹的地方在:如果输入得数据中有前导零时,输出也要注意前导零,还要注意进位。
#include<stdio.h>#include<cstring>using namespace std;char num[100005];int main(){ int t; int sum; int i,j; scanf("%d",&t); while(t--) { sum=1; num[0]='0'; num[1]='0'; scanf("%s",num+2); int n=strlen(num); for(i=0;i<n;i++) //转换为数值存储 num[i]=num[i]-'0'; while(sum%10) //检查10的倍数 { sum=0; j=n-1; num[j]++; //自加1 while(num[j]>=10) //进位 { num[j-1]+=num[j]/10; num[j]%=10; sum+=num[j]; j--; } for(;j>=0;j--) //每个数字相加 sum+=num[j]; } i=1; if(num[i]==0) //检查最高位是否进位 i++; for(;i<n;i++) //直接输出,不要避开前导0 printf("%d",num[i]); printf("\n"); } return 0;}
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