标签:miss ota lap click names log hash panel memory
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19800 Accepted Submission(s):
6781
#include <bits/stdc++.h> using namespace std; int a[3005]; int b[3000*1500+5];//数组较大开在主函数外面较好! bool cmp(int a,int b) { return a>b; } int main() { int n,m; while(~scanf("%d%d",&n,&m)) { int k=0; for(int i=0;i<n;i++) scanf("%d",&a[i]); for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { b[k++]=a[i]+a[j]; } } sort(b,b+k,cmp); for(int i=0;i<m;i++) { if(i!=0) printf(" "); printf("%d",b[i]); } printf("\n"); } return 0; }
#include <iostream> using namespace std; #define NUM_MAX 10003 #define N 3003 int num[N],total[NUM_MAX]; int main(){ int n,m,i,j; while (scanf("%d%d",&n,&m)!=EOF){ memset(total,0,sizeof(total)); for (i=0;i<n;i++) scanf("%d",&num[i]); for (i=0;i<n;i++) for (j=i+1;j<n;j++) total[num[i]+num[j]]++; for (i=NUM_MAX;i>=0;i--){ if (total[i]){ printf("%d",i); m--; total[i]--; i++; if (m==0){ printf("\n"); break; } printf(" "); } } } return 0; }
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<set> #include<map> #include<queue> #include<vector> #include<stack> #include<ctime> #include<cstdlib> #define mem(a,b) memset(a,b,sizeof(a)) #define M 1000005 typedef long long ll; using namespace std; int a[3005],b[10005]; int main() { int n,m,i,j; while(cin>>n>>m) { mem(b,0); int k=0; for(i=0; i<n; i++) cin>>a[i]; for(i=0; i<n; i++) for(j=i+1; j<n; j++) b[a[i]+a[j]]++; //直接把相加的数当作地址,也就是下标 for(i=10000; i>0&&m>0;) { if(!b[i]) {i--;continue;} if(k) cout<<‘ ‘<<i; //因为空格没处理PE了一发 else cout<<i; k=1; b[i]--; //相加可能有相同的,而输出可以有重复的,和北航校赛那题太像了! m--; } cout<<endl; } return 0; }
标签:miss ota lap click names log hash panel memory
原文地址:http://www.cnblogs.com/Roni-i/p/7629191.html