import java.util.*; import java.math.BigInteger; public class Solution { /** * 返回c[n]%1000000007的值 * @param n long长整型 即题目中的n * @return int整型 */ public int Answerforcn (long n) { // write code her BigInteger bi1=new BigInteger("15"); BigInteger bi12=new BigInteger("14"); BigInteger bix=new BigInteger(""+(n-1)); BigInteger bimod=new BigInteger("1000000007"); bi1 = bi1.modPow(bix, bimod).multiply(bi12); int cn=bi1.mod(bimod).intValue(); return cn; } }
通过大整数的modPow()方法直接求出指数幂后取mod,偷懒解法