codeforces Round #320 (Div. 2) C. A Problem about Polyline(数学) D. "Or" Game(暴力,数学)
or quot 数学 Codeforces div round 暴力 Problem
2023-09-14 08:57:55 时间
解题思路:就是求数 n 对应的二进制数中有多少个 1
解题思路:对(strength, i, j)按照strength进行递减排序,从左到右进行遍历,用b[N]表示i和j有关系!
如果发现b[i]或者b[j]有关系了,则跳过这个strength, 否则b[i] =j, b[j] = i
#include iostream #include algorithm #include cstdio using namespace std; struct node{ int x; int i, j; }a[320000]; int b[1000]; bool cmp(node a, node b){ return a.x b.x; int main(){ int x, n; int c = 0; cin n; for(int k=2; k =2*n; ++k){ for(int kk=1; kk ++kk){ cin x; a[c].x = x; a[c].i = k; a[c++].j = kk; sort(a, a+c, cmp); int cnt = 0; for(int i=0; i ++i){ if(!b[a[i].i] !b[a[i].j]){ b[a[i].i] = a[i].j; b[a[i].j] = a[i].i; ++cnt; if(cnt == n) break; for(int i=1; i =2*n; ++i){ if(i!=1) cout " "; cout b[i]; cout endl; return 0; }
解题思路:
我们可以发现这样的一个规律:
(1)首先b一定要小于a,否则无论如何曲线也无法通过(a,b);
(2)设int k=a/b, 如果k为奇数,说明这个点在上图的绿色的线上, 没关系,我们让 k+=1;这样的话一定有(0,0), (a,b)这两点确定的直线的
斜率1/k介于(1/(k-1), 1/(k+1))之间,那么我们可以通过缩小(或者放大)X的值,使得第 k/2 个周期块 斜率为-1的那条边经过(a, b)。此时
的X值就是最小的!
(3)如果(a,b)在第 k/2 个周期块 斜率为-1的那条边上,那么这条边与X轴的交点就是(a+b, 0), 从(0, 0)到(a+b, 0)一共经过了 k/2个周期,
所以 X = (a+b)*1.0/(k/2 * 2)
(4)唉....想的这么明白,容易吗.....
#include iostream #include algorithm #include cstdio #include cmath using namespace std; int main(){ int a, b; cin a b; if(b a) { cout -1 endl; } else { int k = a/b; if(k 1) ++k; printf("%.12lf\n", (a+b)*1.0/k); return 0; }
解题思路:如果某个数a[i]乘以x, 必定会导致a[i]的二进制的长度增大。
首先求出或运算的前缀和后缀,然后对每个a[i]操作如下: a[i]*=x^k(x的k次方); 最后找到a[i]|pref[i-1]|suff[i+1]的最大值!
其实可以优化一处,就是a[i]|pref[i-1]|suff[i+1]的最大值一定对应二进制长度最大的a[i]; 可通过log(a[i])+1求得二进制长度!
#include iostream #include algorithm #include cstdio #include cmath #define N 200010 using namespace std; __int64 a[N]; __int64 pref[N]; __int64 suff[N]; int n, k, x; int main(){ scanf("%d%d%d", n, k, long long maxN = 0; for(int i=1; i ++i) scanf("%I64d", a[i]); long long xk = (long long)(pow((double)x, (double)k) + 0.5); for(int i=1; i ++i){ pref[i] = pref[i-1] | a[i]; suff[n-i+1] = suff[n-i+2] | a[n-i+1]; for(int i=1; i ++i){ long long num = a[i]*xk | pref[i-1] | suff[i+1]; if(maxN num) maxN = num; printf("%I64d\n", maxN); return 0; }
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