Query
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1267 Accepted Submission(s): 391
Your task is to answer next queries:
1) 1 a i c - you should set i-th character in a-th string to c;
2) 2 i - you should output the greatest j such that for all k (i<=k and k<i+j) s1[k] equals s2[k].
Next T blocks contain each test.
The first line of test contains s1.
The second line of test contains s2.
The third line of test contains Q.
Next Q lines of test contain each query:
1) 1 a i c (a is 1 or 2, 0<=i, i<length of a-th string, 'a'<=c, c<='z')
2) 2 i (0<=i, i<l1, i<l2)
All characters in strings are from 'a'..'z' (lowercase latin letters).
Q <= 100000.
l1, l2 <= 1000000.
Then for each query "2 i" output in single line one integer j.
#include<stdio.h>
#include<string.h>
#define lowbit(a) (a&(-a))
#define N 1000005
#define MIN(a,b) (a>b?a:b)
int num[N];
char s[2][N];
int n;
void update(int a,int b)
{
while(a<=n)
{
num[a]+=b;
a+=lowbit(a);
}
}
int sum1(int x)
{
int sum=0;
while(x)
{
sum+=num[x];
x-=lowbit(x);
}
return sum;
}
int sum2(int a,int b)
{
return sum1(b)-sum1(a-1);
}
int main()
{
int t=0;
int cnt=1;
scanf("%d",&t);
while(t--)
{
printf("Case %d:\n",cnt++);
memset(num,0,sizeof(num));
scanf("%s%s",s[0],s[1]);
n=MIN(strlen(s[0]),strlen(s[1]));
for(int i=0;i<n;i++)
update(i+1,s[0][i]==s[1][i] );
int k;
int p,a,b,x;
char c;
scanf("%d",&k);
while(k--)
{
scanf("%d",&p);
if(p==1)
{
scanf("%d%d %c",&a,&b,&c);
a--;
if(s[a][b]!=c&&s[1-a][b]==c)
update(b+1,1);
if(s[a][b]==s[1-a][b]&&s[a][b]!=c)
update(b+1,-1);
s[a][b]=c;
}
if(p==2)
{
scanf("%d",&x);
x++;
int l=x;
int r=n;
while(l<=r)//二分查询；
{
int m=(l+r)/2;
if(sum2(x,m)<(m-x+1))
r=m-1;
else
l=m+1;
}
printf("%d\n",l-x);
}
}
}
return 0;
}
