[第二次训练]Little Sheep and a paper
2013-07-10 23:10
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题目略,题目传送门http://acm.hust.edu.cn/problem.php?cid=1106&pid=7。
解题报告
此题意思不难懂,就是拿一张纸,左右折,上下折,计算出有几条线段是凸出来了。第一次不论怎么折都是0条,然后能找出一个规律。就是记录左右和上下折的次数i,j(不包含第一次),所以ans=(2^i-1)*2^j+(2^j-1)*2^i.代码如下:#include<stdio.h> #include<string.h> char s[1000003]; int main() { long long m,n,x,y,ans; int gz,len,i; scanf("%d",&gz); while(gz--) { scanf("%s",s); len=strlen(s); m=1; n=1; x=0; y=0; for(i=1;i<len;i++) { if(s[i]=='L'||s[i]=='R') { m*=2; m=m%100000009; } if(s[i]=='D'||s[i]=='U') { n*=2; n=n%100000009; } } x=(m-1)*n; y=(n-1)*m; ans=(x+y)%100000009; printf("%lld\n",ans); } return 0; }
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