HDU 1271 整数对 数学题

思路：开始并不会，上网搜的题解。

假设删掉$A$中第$k+1$位

$A = a + b*10^k+c*10^{k+1}$

$B = a + c*10^k$

$N = 2 a + (b + 11 c)*10^k$

所以$N \%10^k = (b + 11c)$，$c = (N\%10^k)/11$ 注意$a<10^k$,但是$2*a$可能产生进位$b$最大只可能是$9$,所以就算a进位了也不会影响c的结果。对于进位的情况分类讨论即可。可能会有重的，譬如5002，需要去重。

http://acm.hdu.edu.cn/showproblem.php?pid=1271

/*********************************************
    Problem : HDU 1271
    Author  : NMfloat
    InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};
int base[11];

void init() {
    base[0] = 1;
    rep(i,1,9) base[i] = base[i-1] * 10;
}

void input() {

}

int a,b,c,N;
int ans[20];
int tot ;

void solve() {
    //N = 2*a+(b + 11c)*10^k;
    tot = 0;
    rep(i,0,9) {
        if(N >= base[i]) {
            if((N % base[i])&1)continue;
            int bc = N / base[i];
            c = bc / 11;
            b = bc % 11; 
            a = (N % base[i]) / 2;
            if(b == 10) {
                if(i == 0) continue;
                b --;
                a += base[i] / 2;
                ans[++tot] = a + (b + 10*c) * base[i];
                //printf("a : %d b : %d c : %d\n",a,b,c);
                //printf("%d\n",ans[tot]);
                continue;
            }
            ans[++tot] = a + (b + 10*c) * base[i];
            //printf("a : %d b : %d c : %d\n",a,b,c);
            //printf("%d\n",ans[tot]);
            if(i > 0 && b >= 1) {
                b --;
                if(b == 0 && c == 0) continue;
                a += base[i] / 2;
                ans[++tot] = a + (b + 10*c) * base[i];
                //printf("a : %d b : %d c : %d\n",a,b,c);
                //printf("%d\n",ans[tot]);
            }
        }
    }
    if(tot == 0) {
        puts("No solution.");
    }
    else {
        sort(ans+1,ans+tot+1);
        printf("%d",ans[1]);
        rep(i,2,tot) {
            if(ans[i] != ans[i-1]) printf(" %d",ans[i]);
        }
        puts("");
    }
}

int main(void) {
    //freopen("a.in","r",stdin);
    init();
    while(scanf("%d",&N),N) {
        input();
        solve();
    }
    return 0;
}