12 2012 档案

摘要:#include#includeusing namespace std;#define N 1000000010#define M 1000010__int64 sum;//long long sum;int a[M],L[M/2+1],R[M/2+1];void merger(int beg,in... 阅读全文
posted @ 2012-12-30 15:24 YaLing 阅读(179) 评论(1) 推荐(0)
摘要://本题主要运用了树组数组的知识,学了后就会做了#include#includeconst int N=1000010;int n,c[N];//该函数功能是求出n二进制中最右边0的个数的2次幂,也等于c[n]包含的元素个数num[n-lowbit(n)+1]+...+num[n]int lowbi... 阅读全文
posted @ 2012-12-28 22:04 YaLing 阅读(183) 评论(0) 推荐(0)
摘要:#includeint c[1000010];int main(){// freopen("in.txt","r",stdin); int M,N,a,b; scanf("%d%d",&M,&N); c[0]=0; for(int i=1;i<=M;i++) { scanf("%d",&c[i])... 阅读全文
posted @ 2012-12-28 19:06 YaLing 阅读(154) 评论(0) 推荐(0)