MUTC2013 J-I-number-hdu4608

I-number

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 195    Accepted Submission(s): 75

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 10⁵.

 

Output

Output the I-number of x for each query.

 

Sample Input

1202

 

Sample Output

208

 

Source

2013 Multi-University Training Contest 1

 

Recommend

liuyiding

 

模拟。。即可

/** head-file **/#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <vector>#include <queue>#include <stack>#include <list>#include <set>#include <map>#include <algorithm>/** define-for **/#define REP(i, n) for (int i=0;i<int(n);++i)#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)#define REP_1(i, n) for (int i=1;i<=int(n);++i)#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)#define REP_N(i, n) for (i=0;i<int(n);++i)#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)#define REP_1_N(i, n) for (i=1;i<=int(n);++i)#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)/** define-useful **/#define clr(x,a) memset(x,a,sizeof(x))#define sz(x) int(x.size())#define see(x) cerr<<#x<<" "<<x<<endl#define se(x) cerr<<" "<<x#define pb push_back#define mp make_pair/** test **/#define Display(A, n, m) {                      \    REP(i, n){                                  \        REP(j, m) cout << A[i][j] << " ";       \        cout << endl;                           \    }                                           \}#define Display_1(A, n, m) {                    \    REP_1(i, n){                                \        REP_1(j, m) cout << A[i][j] << " ";     \        cout << endl;                           \    }                                           \}using namespace std;/** typedef **/typedef long long LL;/** Add - On **/const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };const int MOD = 1000000007;const int INF = 0x3f3f3f3f;const long long INFF = 1LL << 60;const double EPS = 1e-9;const double OO = 1e15;const double PI = acos(-1.0); //M_PI;const int maxn=111111;char str[maxn];int a[maxn];int n,sum;void digadd(int p){    a[p]++;    sum++;    int ts=p;    while (a[ts]>=10)    {        a[ts]-=10;        sum-=10;        a[ts+1]++;        sum++;        ts++;        if (ts>=n) n++;    }}int main(){    int T;    cin>>T;    while (T--)    {        sum=0;        clr(a,0);        cin>>str;        n=strlen(str);        REP(i,n)        {            a[i]=str[n-i-1]-'0';            sum+=a[i];        }        do        {            digadd(0);        }while (sum%10!=0);        REP(i,n) cout<<a[n-i-1];        cout<<endl;    }    return 0;}