题意
1<=a,b,c<=1e100000
思路：
1 #include<bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 typedef unsigned int uint;
5 typedef unsigned long long ull;
6 typedef pair<int,int> PII;
7 typedef pair<ll,ll> Pll;
8 typedef vector<int> VI;
9 typedef vector<PII> VII;
10 #define N 1100000
11 #define M 4100000
12 #define fi first
13 #define se second
14 #define MP make_pair
15 #define pi acos(-1)
16 #define mem(a,b) memset(a,b,sizeof(a))
17 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
18 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
19 #define lowbit(x) x&(-x)
20 #define Rand (rand()*(1<<16)+rand())
21 #define id(x) ((x)<=B?(x):m-n/(x)+1)
22 #define ls p<<1
23 #define rs p<<1|1
24
25 const ll MOD=1e9+7,inv2=(MOD+1)/2;
26 double eps=1e-6;
27 int INF=1e9;
28
29 char a[N],b[N],c[N];
30 int A[N],B[N],C[N],t[N],flag,la,lb,lc;
31
32 int read()
33 {
34 int v=0,f=1;
35 char c=getchar();
36 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
37 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
38 return v*f;
39 }
40
41 void solve1()
42 {
43 int na=0,nc=0;
44 rep(i,1,1e6+1) A[i]=B[i]=C[i]=t[i]=0;
45 rep(i,1,1e6-la) A[++na]=0;
46 per(i,la,1) A[++na]=a[i]-'0';
47 rep(i,1,1e6-lc) C[++nc]=0;
48 per(i,lc,1) C[++nc]=c[i]-'0';
49 int p=1;
50 per(i,1e6,1)
51 {
52 if(C[i]>A[i]) break;
53 if(C[i]<A[i]){p=0; break;}
54 }
55 if(!p) return;
56 rep(i,1,1e6)
57 {
58 t[i]=t[i]+C[i]-A[i];
59 if(t[i]<0)
60 {
61 t[i+1]--;
62 t[i]+=10;
63 }
64 }
65 p=1;
66 int k=1e6;
67 while(k>1&&t[k]==0) k--;
68 per(i,k/2,1) swap(t[i],t[k-i+1]);
69 rep(i,1,lb)
70 if((b[i]-'0')!=t[i]){p=0; break;}
71 if(p)
72 {
73 flag=1;
74 printf("%d %d %d\n",1000000-la,k-lb,1000000-lc);
75 }
76 }
77
78 void solve2()
79 {
80 if(flag) return;
81 rep(i,1,1e6+1) A[i]=B[i]=C[i]=t[i]=0;
82 int nb=0,nc=0;
83 rep(i,1,1e6-lb) B[++nb]=0;
84 per(i,lb,1) B[++nb]=b[i]-'0';
85 rep(i,1,1e6-lc) C[++nc]=0;
86 per(i,lc,1) C[++nc]=c[i]-'0';
87 int p=1;
88 per(i,1e6,1)
89 {
90 if(C[i]>B[i]) break;
91 if(C[i]<B[i]){p=0; break;}
92 }
93 if(!p) return;
94 rep(i,1,1e6)
95 {
96 t[i]=t[i]+C[i]-B[i];
97 if(t[i]<0)
98 {
99 t[i+1]--;
100 t[i]+=10;
101 }
102 }
103 p=1;
104 int k=1e6;
105 while(k>1&&t[k]==0) k--;
106 per(i,k/2,1) swap(t[i],t[k-i+1]);
107 rep(i,1,la)
108 if((a[i]-'0')!=t[i]){p=0; break;}
109 if(p)
110 {
111 flag=1;
112 printf("%d %d %d\n",k-la,1000000-lb,1000000-lc);
113 }
114 }
115
116 void solve3()
117 {
118 if(flag) return;
119 rep(i,1,1e6+1) A[i]=B[i]=C[i]=t[i]=0;
120 int na=0,nb=0,nc=0;
121 rep(i,1,1e6-1-la) A[++na]=0;
122 per(i,la,1) A[++na]=a[i]-'0';
123 rep(i,1,1e6-1-lb) B[++nb]=0;
124 per(i,lb,1) B[++nb]=b[i]-'0';
125 rep(i,1,1e6-lc) C[++nc]=0;
126 per(i,lc,1) C[++nc]=c[i]-'0';
127 rep(i,1,1e6)
128 {
129 t[i]=t[i]+A[i]+B[i];
130 if(t[i]>9)
131 {
132 t[i]-=10;
133 t[i+1]++;
134 }
135 }
136 int p=1;
137 rep(i,1,1e6+1)
138 if(t[i]!=c[i]){p=0; break;}
139 if(p)
140 {
141 flag=1;
142 printf("%d %d %d\n",1000000-1-la,1000000-1-lb,1000000-lc);
143 }
144 }
145
146 int main()
147 {
148 //freopen("1.in","r",stdin);
149 int cas;
150 scanf("%d",&cas);
151 while(cas--)
152 {
153 scanf("%s",a+1);
154 scanf("%s",b+1);
155 scanf("%s",c+1);
156 la=strlen(a+1),lb=strlen(b+1),lc=strlen(c+1);
157 //printf("la=%d lb=%d lc=%d\n",la,lb,lc);
158 flag=0;
159 solve1();
160 solve2();
161 solve3();
162 if(!flag) printf("-1\n");
163 }
164
165 return 0;
166 }