POJ3785:The Next Permutation——全排列问题


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.

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.

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. If there 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

1 123
2 279134399742
3 987
Sample Output

1 132
2 279134423799


输入数组时用scanf()而不用gets()。原因是scanf()遇到空格时认为输入结束,而gets()会输入空格:样例显然是以空格而非 结束标志来判断的。所以,乖乖scanf()吧。



char num[1010];
using namespace std;
int main()
	int p, n, cache;
	scanf("%d", &p);
	while (p--)
		memset(num, 0, sizeof(num));
		cache = n = 0;
		scanf("%d", &cache);
		scanf("%s", num);
		printf("%d ", cache);
		if (next_permutation(num, num + strlen(num)))
		else printf("BIGGESTn");
	return 0;




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