1044. Shopping in Mars (25)
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采用O(n^2)的做法,会有两个case超时。可以在金额刚好等于m的时候,省略再次从start+1 到end的和。优化后的代码如下:
#include<cstdio>#include<vector>using namespace std;#define N 100000int diam[N];struct pir{ int s,e;};int main(){ //freopen("1.txt","r",stdin); int n,m,i,j,sum,start,end,min; vector<pir> pairs; scanf("%d %d",&n,&m); for(i=0;i<n;i++){ scanf("%d",&diam[i]); } min = 999999; bool flag = false; int jmp =0; sum=0; for(i=0;i<n;i++){ start = i; for(j=start+jmp;j<n;j++){ sum+=diam[j]; if(sum>=m){ end = j; break; } } if(sum==m){ printf("%d-%d\n",start+1,end+1); flag=true; jmp=end-start; sum-=diam[start]; }else if(sum>m){ if(sum<min){ pir p; p.s=start; p.e = end; pairs.clear(); pairs.push_back(p); min = sum; }else if(sum==min){ pir p; p.s=start; p.e = end; pairs.push_back(p); } sum=0; jmp=0; } } if(!flag){ for(i=0;i<pairs.size();i++){ printf("%d-%d\n",pairs[i].s+1,pairs[i].e+1); } }}
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- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)
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- 1044. Shopping in Mars (25)
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