国内php开发的电商网站有哪些,网站建站的职位,设计一个网站首页方案,建筑工程网络副业http://www.lydsy.com/JudgeOnline/problem.php?id1679 水题没啥好说的。。自己用笔画画就懂了 将点排序#xff0c;然后每一次的点到后边点的声音距离和(n-i)*(a[i1]-a[i])之前同样操作所得的的sum 然后答案就是累加后2 #include cstdio
#include cstring
#…http://www.lydsy.com/JudgeOnline/problem.php?id1679 水题没啥好说的。。自己用笔画画就懂了 将点排序然后每一次的点到后边点的声音距离和(n-i)*(a[i1]-a[i])之前同样操作所得的的sum 然后答案就是累加后×2 #include cstdio
#include cstring
#include cmath
#include string
#include iostream
#include algorithm
using namespace std;
#define rep(i, n) for(int i0; i(n); i)
#define for1(i,a,n) for(int i(a);i(n);i)
#define for2(i,a,n) for(int i(a);i(n);i)
#define for3(i,a,n) for(int i(a);i(n);--i)
#define for4(i,a,n) for(int i(a);i(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) agetint()
#define print(a) printf(%d, a)
#define dbg(x) cout #x x endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout a[aaa][bbb]; cout endl; }
inline const int getint() { int r0, k1; char cgetchar(); for(; c0||c9; cgetchar()) if(c-) k-1; for(; c0c9; cgetchar()) rr*10c-0; return k*r; }
inline const int max(const int a, const int b) { return ab?a:b; }
inline const int min(const int a, const int b) { return ab?a:b; }
const int N10005;
int n;
long long ans, sum[N], a[N];
int main() {read(n);for1(i, 1, n) scanf(%lld, a[i]);sort(a1, a1n);for3(i, n, 1) { sum[i](n-i)*(a[i1]-a[i])sum[i1]; anssum[i]; }printf(%lld, ans1);return 0;
}Description Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise. FJs N cows (1 N 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions. 约翰的邻居鲍勃控告约翰家的牛们太会叫 约翰的N(1≤N≤10000)只牛在一维的草场上的不同地点吃着 草她们都是些爱说闲话的奶牛每一只同时与其他N-1只牛聊着天一个对话的进行需要两只牛都按照和她们间距离等大的音量吼叫因此草场上存在着 NN-1/2个声音 请计算这些音量的和 Input * Line 1: N * Lines 2..N1: The location of each cow (in the range 0..1,000,000,000). 第1行输入N接下来输入N个整数表示一只牛所在的位置 Output * Line 1: A single integer, the total volume of all the MOOs. 一个整数表示总音量 Sample Input 5 1 5 3 2 4 INPUT DETAILS: There are five cows at locations 1, 5, 3, 2, and 4. Sample Output 40 OUTPUT DETAILS: Cow at 1 contributes 123410, cow at 5 contributes 432110, cow at 3 contributes 21126, cow at 2 contributes 11237, and cow at 4 contributes 32117. The total volume is (1010677) 40. HINT Source Silver
