1 条题解
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0
C :
#include<stdio.h> #include<string.h> #include<math.h> int main(){ int N; while(scanf("%d",&N)!=EOF){ int i,t,s1=0,s2=0; for(i=0;i<2*N;i++){ scanf("%d",&t); if(i%2) s1+=t; else s2+=t;} if(s1!=s2) printf("YES\n"); else printf("NO\n");} return 0;}
C++ :
#include<bits/stdc++.h> using namespace std; int n,x,ou,ji; int main() { cin>>n; for(int i=0;i<2*n;i++) { cin>>x; if(i%2==0) ou+=x; else ji+=x; } if(ji!=ou) cout<<"YES"; else cout<<"NO"; return 0; }
Pascal :
var a:array[1..201] of longint; x,maxi,i,s1,s2:longint; begin readln(x); maxi:=1; for i:=1 to x*2 do begin read(a[i]); if i mod 2=1 then inc(s1,a[i]) else inc(s2,a[i]); if a[i]>a[maxi] then maxi:=i; end; if maxi mod 2=1 then if s1<=s2 then writeln('NO') else writeln('YES') else if s2<=s1 then writeln('NO') else writeln('YES'); end.
- 1
信息
- ID
- 3609
- 时间
- 1000ms
- 内存
- 64MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者