水题水坑，虐心到哭

   今天下午参加一场训练比赛，碰到一没话说的题目，本以为这是我会做的，想想本以为自己会做，结果，唉 一言难尽。突然发现今天好累但是正因为我学的累，才有挑战的动力，加油晚安。

题目如下

Easy Task

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Time Limit: 2 Seconds                                    Memory Limit: 65536 KB                            

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You are given n integers. Your task is very easy. You should find the maximum integera and the minimum integer b among these n integers. And then you should replace botha and b with a-b. Your task will not be finished unless all the integers are equal.

Now the problem come, you want to know whether you can finish you task. And if you can finish the task, you want to know the final result.

Input

The first line of the input contain an integer T(T≤ 20) indicates the number of test cases.

Then T cases come. Each case consists of two lines. The first line is an integern(2≤ n≤ 10) as the problem described. The second line containsn integers, all of them are no less than -100000 and no more than 100000.

Output

For each case you should print one line. If you can finish your task, you should print one of then integers. Otherwise, you should print "Nooooooo!"(without quotes).

Sample Input

231 2 325 5

Sample Output

25

一目了然的解决方法，先贴上自己能完美输出样例，结果一直都是wa

#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int nn[25];int main(){ int T,n,a,b,i,m,fg=0; while(scanf("%d",&T)!=EOF) { while(T--) { fg=0; scanf("%d",&n); // if(n<2 n="">10)fg=1; for( i=0;i<n i="" scanf="" d="" nn="" i="" if="" nn="" i="">100000||nn[i]<-100000) { fg=1; } } if(fg)printf("Nooooooo!\n"); int k=i; if(!fg) { for(;;) { sort(nn,nn+k); a=nn[k-1]; b=nn[0]; m=a-b; if(m==0) { printf("%d\n",a); break; } else { for(i=0;i<n;i++) { if(nn[i]==a||nn[i]==b) { nn[i]=m; } } } } } } } return 0;}</n></2></algorithm></cmath></cstdio>

这只有机会在解决了，

现在附上一个正确的

 //ZOJ Problem Set - 3844 Easy Task#include<cstdio>#include<cmath>#include<algorithm>#define inff 0x3fffffffusing namespace std;int nn[20];int main(){ int T,n,i,id1,id2,m,a,b; while(scanf("%d",&T)!=EOF) { while(T--) { scanf("%d",&n); for(i=0;i<n;i++) scanf("%d",&nn[i]); while(1) { a=-inff,b=inff; //各种虐心，我手抽竟然把这里改成 inta=-inff,b=inff;这就导致我输出了错误的东西。 for(i=0;i<n i="" if="" nn="" i="">a) { a=nn[i]; id1=i; } if(nn[i]<b) { b=nn[i]; id2=i; } } // printf("%d %d\n",a,b); m=a-b; if(m==0) break; nn[id1]=m; nn[id2]=m; } printf("%d\n",a); } } return 0;}</n></algorithm></cmath></cstdio>

 同时在网上我找到一个奇葩题解，复杂到新手完全看不懂

#include <iostream>#include <stdio h="">#include <algorithm>#include <string h="">#include <stdlib h="">#include <cmath>#include <iomanip>#include <vector>#include <set>#include <map>#include <stack>#include <queue>#include <cctype>using namespace std;#define ll long longconst int mod=100000*10+1;const int inf=0x3f3f3f3f;const int maxn=12;int num[maxn];int n;struct Node{ int a[maxn];};vector<node>vc[mod+10];int main(){ int t; cin>>t; while(t--) { for(int i=0;i<=mod;i++) vc[i].clear(); Node tp; cin>>n; int sum=0; for(int i=0;i<n i="" cin="">>num[i]; sum+=num[i]; } tp.a[n]=sum; sort(num,num+n); if(num[0]==num[n-1]) { cout<<num[0]<<endl; continue; } for(int i=0;i<n;i++) tp.a[i]=num[i]; vc[(tp.a[n]+mod)%mod].push_back(tp); int ans=inf; while(1) { int cha=num[n-1]-num[0]; sum=sum-num[n-1]-num[0]+2*cha;//新的sum num[0]=cha,num[n-1]=cha; sort(num,num+n); if(num[0]==num[n-1]) { ans=num[0]; break; } tp.a[n]=sum; for(int i=0;i<n;i++) tp.a[i]=num[i]; bool ok=0; int idx=(tp.a[n]+mod)%mod;//去相应的索引中去找看是否有相同的状态 for(int i=0;i<vc[idx].size();i++) { int j; for(j=0;j<n;j++) { if(tp.a[j]!=vc[idx][i].a[j]) break; } if(j==n) { ok=1;//找到了！ break; } } if(ok==1) break; vc[idx].push_back(tp); } if(ans!=inf) cout<<ans<<endl; else cout<<"Nooooooo!"<<endl; } return 0;}</n></node></cctype></queue></stack></map></set></vector></iomanip></cmath></stdlib></string></algorithm></stdio></iostream>