poj2231 2010.8.1
2016-02-13 18:08
405 查看
Moo Volume
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 11299 Accepted: 3177
Description
Farmer John has received a noise complaintfrom his neighbor, Farmer Bob, stating that his cows are making too much noise.
FJ's N cows (1 <= N <= 10,000) allgraze at various locations on a long one-dimensional pasture. The cows are verychatty animals. Every pair of cows simultaneously carries on a conversation (soevery cow is simultaneously MOOing at all of the N-1 other cows).
When cow iMOOs at cow j, the volume of this MOO must be equal to the distance between iand j, in order for j to be able to hear the MOO at all. Please help FJ computethe total volume of sound being generated by all N*(N-1) simultaneous MOOingsessions.
Input
* Line 1: N
* Lines 2..N+1: The location of each cow(in the range 0..1,000,000,000).
Output
There are five cows at locations 1, 5, 3,2, and 4.
Sample Input
5
1
5
3
2
4
Sample Output
40
Hint
INPUT DETAILS:
There are five cows at locations 1, 5, 3,2, and 4.
OUTPUT DETAILS:
Cow at 1 contributes 1+2+3+4=10, cow at 5contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is(10+10+6+7+7) = 40.
Source
USACO 2005 January Silver
分析:
水题,但是。。。
刚开始觉得太水了,暴力就能过。。。但是暴力是不行的。。后来搜了解题报告,发现,是一道递推+数学的题。。。。
1.先把牛的坐标从小到大排序;
2.牛i 和牛i-1 之间的这段距离,被用了的次数=这个距离的左边的牛数*这个距离右边的牛数;
3.把这些都加起来,*2,就是结果了。
很强,我都米想到。。。。一定要抓住本质。。。
Source Code
Problem: 2231 User: creamxcream
Memory: 216K Time: 16MS
Language: C++ Result: Accepted
Source Code
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 11299 Accepted: 3177
Description
Farmer John has received a noise complaintfrom his neighbor, Farmer Bob, stating that his cows are making too much noise.
FJ's N cows (1 <= N <= 10,000) allgraze at various locations on a long one-dimensional pasture. The cows are verychatty animals. Every pair of cows simultaneously carries on a conversation (soevery cow is simultaneously MOOing at all of the N-1 other cows).
When cow iMOOs at cow j, the volume of this MOO must be equal to the distance between iand j, in order for j to be able to hear the MOO at all. Please help FJ computethe total volume of sound being generated by all N*(N-1) simultaneous MOOingsessions.
Input
* Line 1: N
* Lines 2..N+1: The location of each cow(in the range 0..1,000,000,000).
Output
There are five cows at locations 1, 5, 3,2, and 4.
Sample Input
5
1
5
3
2
4
Sample Output
40
Hint
INPUT DETAILS:
There are five cows at locations 1, 5, 3,2, and 4.
OUTPUT DETAILS:
Cow at 1 contributes 1+2+3+4=10, cow at 5contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is(10+10+6+7+7) = 40.
Source
USACO 2005 January Silver
分析:
水题,但是。。。
刚开始觉得太水了,暴力就能过。。。但是暴力是不行的。。后来搜了解题报告,发现,是一道递推+数学的题。。。。
1.先把牛的坐标从小到大排序;
2.牛i 和牛i-1 之间的这段距离,被用了的次数=这个距离的左边的牛数*这个距离右边的牛数;
3.把这些都加起来,*2,就是结果了。
很强,我都米想到。。。。一定要抓住本质。。。
Source Code
Problem: 2231 User: creamxcream
Memory: 216K Time: 16MS
Language: C++ Result: Accepted
Source Code
#include <iostream> #include <algorithm> using namespace std; #define MAXN 10010 __int64 position[MAXN]; int n; bool cmp(__int64 a,__int64 b) { return (a<b); } int main() { scanf("%d",&n); memset(position,0,sizeof(position)); for(int i=1;i<=n;i++) scanf("%I64d",&position[i]); sort(position+1,position+n+1,cmp); __int64 ans=0; for(int i=2;i<=n;i++) ans+=(position[i]-position[i-1])*(i-1)*(n-i+1); printf("%I64d\n",ans*2); return 0; }
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