A correct expression of the form a+b=c was written; a, b and c are non-negative integers without leading zeros. In this expression, the plus and equally signs were lost. The task is to restore the expression. In other words, one character '+' and one character '=' should be inserted into given sequence of digits so that:
- character'+' is placed on the left of character '=',
- characters '+' and '=' split the sequence into three non-empty subsequences consisting of digits (let's call the left part a, the middle part — b and the right part — c),
- all the three parts a, b and c do not contain leading zeros,
- it is true that a+b=c.
It is guaranteed that in given tests answer always exists.
The first line contains a non-empty string consisting of digits. The length of the string does not exceed 106.
Output the restored expression. If there are several solutions, you can print any of them.
Note that the answer at first should contain two terms (divided with symbol '+'), and then the result of their addition, before which symbol'=' should be.
Do not separate numbers and operation signs with spaces. Strictly follow the output format given in the examples.
If you remove symbol '+' and symbol '=' from answer string you should get a string, same as string from the input data.
12345168
123+45=168
099
0+9=9
199100
1+99=100
123123123456456456579579579
123123123+456456456=579579579
大意：其实看样例就行了
题解：A+B=C A是x位数，B是y位数，C是z位数， x 和 y 中至少有一个和 z 相等或比 z 小 1。
有了这个条件，就可以O(L)枚举+的位置，然后O(1)找出=可能在的位置，关键就是怎么检验值相等？
RK hash给了我启示，长度为len的序列的前缀1……m的hash值可以用一个MOD进制数表示，想要知道x……y的hash值：
Hash[x,y]=Hash[1,y]-Hash[1,x-1]*10^(y-x+1)
多取几个MOD来检验就基本可以保证正确性。
/*
Welcome Hacking
Wish You High Rating
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<string>
using namespace std;
int read(){
int xx=0,ff=1;char ch=getchar();
while(ch>'9'||ch<'0'){if(ch=='-')ff=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){xx=(xx<<3)+(xx<<1)+ch-'0';ch=getchar();}
return xx*ff;
}
const int maxn=1000010;
char s[maxn];
int len;
int MOD[5]={0,1000000007,19797571,73715923,92431371};
//int MOD[5]={0,1000000,1000000,1000000,1000000};
int Hash[5][maxn];
int mypow(int a,int p,int mod){
int re=1;
while(p){
if(p&1)
re=1LL*re*a%mod;
p>>=1;
a=1LL*a*a%mod;
}
return re;
}
int get_Hash(int xx,int yy,int mod){
//printf("%d %d %d %d %d
",xx,yy,Hash[mod][yy],Hash[mod][xx-1],mypow(10,yy-xx+1,MOD[mod]));
return ((Hash[mod][yy]-1LL*Hash[mod][xx-1]*mypow(10,yy-xx+1,MOD[mod]))%MOD[mod]+MOD[mod])%MOD[mod];
}
bool check(int x,int y){
if(s[1]=='0'&&x!=1)
return 0;
if(s[x+1]=='0'&&y!=x+1)
return 0;
if(s[y+1]=='0'&&len!=y+1)
return 0;
for(int i=1;i<=4;i++)
if((1LL*get_Hash(1,x,i)+get_Hash(x+1,y,i))%MOD[i]!=get_Hash(y+1,len,i))
return 0;
return 1;
}
int main(){
//freopen("in","r",stdin);
//freopen("out","w",stdout);
gets(s+1);
len=strlen(s+1);
for(int i=1;i<=4;i++)
for(int j=1;j<=len;j++)
Hash[i][j]=(1LL*Hash[i][j-1]*10+s[j]-'0')%MOD[i];
/*for(int i=1;i<=4;i++){
for(int j=0;j<=len;j++)
printf("%d ",Hash[i][j]);
puts("");
}*/
int i,j;
for(i=1;i<=len/2-(len%2==0);i++){
j=len-i;
if(j/2+(j%2==1)>i){
j=i+(j/2);
if(check(i,j))
break;
}
else{
j=len-i;
if(i==j)
break;
if(check(i,j))
break;
j=len-i-1;
if(check(i,j))
break;
}
}
for(int k=1;k<=len;k++){
printf("%c",s[k]);
if(k==i)
printf("+");
else if(k==j)
printf("=");
}
puts("");
return 0;
}
细节很多很多，交了五次。
要注意前缀 0 的问题