codeforces 509c

题意：

给出n个数字，这些数字数由另外n个数字各个位数的和得来的。求另外个数，要尽量小，而且要递增。

题解：

从最小位往最大位贪心，如果可以在做到最小位大于上一个数其他位与上一个数相等最好，不行就再往前面试，全部都不行就要位数比前面的大，具体参见：http://blog.csdn.net/ck_boss/article/details/43573911

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <stack>
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define PF(x) cout << "debug: " << x << " ";
#define EL cout << endl;
#define PC(x) puts(x);
typedef long long ll;
#define CLR(x, v) sizeof (x, v, sizeof(x))
using namespace std;
const int INF = 0x5f5f5f5f;
const int maxn = 350;
const int mod = 1e9 + 7;
int pos[maxn],num[maxn][maxn],b[maxn],n;
int main()
{
 //   freopen("in.txt","r",stdin);
    cin>>n;
    for(int i = 1;i <= n;i++)
        scanf("%d",&b[i]);
    int last = b[1];
    while(b[1] >= 9) num[1][++pos[1]] = 9,b[1] -= 9;
    if(b[1] > 0)
    num[1][++pos[1]] = b[1];
    bool fg = false;
    b[1] = last;
    for(int i = 2;i <= n;i++){
        fg = false;
        last = b[i-1];
        for(int j = 1;fg == false;j++){
            last -= num[i-1][j-1];
            for(int k = num[i-1][j] + 1;k <= 9&&fg == false;k++){
                int add = k-num[i-1][j];
                if(last + add <= b[i]&&last + add +9*(j-1) >= b[i]){
                    fg = true;
                    int tn = 0;
                    pos[i] = max(pos[i-1],j);
                    int val = b[i] - last - add;
                    while(val > 0){
                        if(val >= 9){ num[i][++tn] = 9,val -= 9;}
                        else if(val > 0){num[i][++tn] = val,val = 0;}
                    }
                    num[i][j] = k;
                    for(int t = j+1;t <= pos[i];t++)
                        num[i][t] = num[i-1][t];
                }

            }

        }
    }
    for(int i = 1;i <= n;i++){
        for(int j = pos[i];j >= 1;j--)
            printf("%d",num[i][j]);
        printf("
");
    }
    return 0;
}