转圈踢人问题
https://www.cnblogs.com/lanxuezaipiao/p/3339603.html
有N個(gè)人圍一圈依次報(bào)數(shù),數(shù)到3的倍數(shù)的人出列,問(wèn)當(dāng)只剩一個(gè)人時(shí)他原來(lái)的位子在哪里?
解答:經(jīng)典的轉(zhuǎn)圈踢人問(wèn)題,好吧專(zhuān)業(yè)一點(diǎn),約瑟夫環(huán)問(wèn)題,相信大家都會(huì),下面給我的code:
int main() {int N, i, j;printf("Please enter the number of people(N): ");scanf("%d", &N);int *pArray = (int *)malloc(sizeof(int) * N);int count = 0;// 這里編號(hào)為0 ~ N - 1for(i = 0; i < N; i++){pArray[i] = i;}for(i = 0, j = 0; i < N; i = (i + 1) % N){if(pArray[i] != -1){j++;if(j % 3 == 0){pArray[i] = -1;count++;if(count == N){printf("The last people is %d\n", i);break;}}}}return 0; }好吧,我承認(rèn)我的算法很臃腫,完全是模擬了整個(gè)游戲過(guò)程,時(shí)間復(fù)雜度為O(mn),這里m=3,網(wǎng)上有個(gè)大牛給出了歸納數(shù)學(xué)的方法,具體方法如下:
創(chuàng)作挑戰(zhàn)賽新人創(chuàng)作獎(jiǎng)勵(lì)來(lái)咯,堅(jiān)持創(chuàng)作打卡瓜分現(xiàn)金大獎(jiǎng)總結(jié)
- 上一篇: Java这个地方,怎么让他不弹出dial
- 下一篇: string类的基本实现