杭电ACM HDU 3283 The Next Permutation

The Next Permutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 618    Accepted Submission(s): 439

Problem Description

For this problem, you will write a program that takes a (possibly long) string of decimal digits, and outputs the permutation of those decimal digits that has the next larger value (as a decimal number)
than the input number. For example:
123 -> 132
279134399742 -> 279134423799
It is possible that no permutation of the input digits has a larger value. For example, 987.

 

 

Input

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set is a single line that contains the data set number, followed by a space, followed by up to 80 decimal digits which is the input value.

 

 

Output

For each data set there is one line of output. If there is no larger permutation of the input digits, the output should be the data set number followed by a single space, followed by the string BIGGEST. Ifthere is a solution, the output should be the data set number, a single space and the next larger
permutation of the input digits.

 

 

Sample Input

3 1 123 2 279134399742 3 987

 

 

Sample Output

1 132 2 279134423799 3 BIGGEST

 

 

Source

2009 Greater New York Regional

 

 

Recommend

zhuweicong

 

 

#include 
#include 
#include 
int cmp(const void *a,const void *b){
    return *(char *)a-*(char *)b;
}
int main(){
    int t,n,len;
    char str[100];
	//freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        scanf("%s",str);
        len=strlen(str);
        int i,j,flag=0;
        for(i=len-1;i>0;i--){
            qsort(str+i,len-i,sizeof(char),cmp);
            for(j=i;j