SPOJ Problem 5:The Next Palindrome

题目大意：根据给出的数字串求较大的最小回文串

    简单题，可以从中间开始比较，如果左边大于右边的话则终止，若右边大于左边的话则中间+1，然后向左边推。输出时按左边复制一遍。

#include<cstdio>
#include<cstring>
int t,i,j,l,r,n,pd,q,mid;
char s[1000005];
int main(){
    scanf("%d",&t);
    while(t--){
        pd=q=l=r=0;
        scanf("%s",s);
        n=strlen(s);
        if(n%2==0)l=1;
        mid=n/2;
        while(l<=mid&&pd==0){
            if (s[mid-l]==s[mid+r]){l++;r++;}else
            if (s[mid-l]<s[mid+r])break;else
            if (s[mid-l]>s[mid+r])pd=1;
        }
        if (pd==0){
            l=1;
            if(n%2)l=0;
            s[mid-l]++;
            while(l<=mid&&s[mid-l]==58){
                s[mid-l]=48;
                if(l<mid){
                    s[mid-l-1]++;
                }
                l++;
            }
            if (l>mid)q=1;
        }
        if (q){printf("1");for (i=1;i<n;i++)
            printf("0");printf("1\n");
        }
        else {
            for(i=0;i<mid;i++)
                printf("%c",s[i]);
            for (n%2==0?i=mid-1:i=mid;i>=0;i--)
                printf("%c",s[i]);
            printf("\n");
        }
    }
}