HDU 1029 Ignatius and the Princess IV
2016-10-20 19:30
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HDU 1029 Ignatius and the Princess IV
传送门题意
求众数
思路
1.桶
2.像个算法的数学的脑经急转弯的算法
从头到尾扫一遍,如果出现了两个不同的数,就把它们一起划掉,重复之。最后一定会剩下一个数(或者一堆值一样的数),即为众数。
代码
#include<cstdio> using namespace std; int main(){ freopen("BBB.in","r",stdin); int n; while(~scanf("%d",&n)){ int a=-1;int cnt=0; while(n--){ int x; scanf("%d",&x); if(cnt<=0)a=x,cnt=0; if(a==x)cnt++; else cnt--; } printf("%d\n",a); } return 0; }
后记
和@DEL做的DP的第二题 第一篇markdown
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