HDU 4608 I-number(模拟)
I-number
Time Limit: 5000ms Memory limit: 65536K 有疑问?点这里^_^
题目描写叙述
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.
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.
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.
演示样例输入
1 202
演示样例输出
208
提示
来源
2013 Multi-University Training Contest 1
我不得不说这道题真的非常坑,我要是不搜题解预计是毁掉了 本来题意非常easy,就是给定一个x(因为可能非常大用字符串模拟),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居然是能够有前导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; }