【openjudge 1.11 07】和为给定数


我看到这个的第一反应就是爆搜,然后写了一个没加$No$的4分程序

#include <bits/stdc++.h>
#include <algorithm>
#define it int
#define fr for
#define il inline
using namespace std;
it book[100020],t[100020];
it spe[50010];
it n,m,bz=0;
it main();
it find();
it max();
it min();
il it min(it a,it b){
    if(a>b){
        return b;
    }
    else{
        return a;
    }
}
il it max(it a,it b){
    if(a>b){
        return a;
    }
    else{
        return b;
    }
}
void print(it cho,it now){
    if(bz==0){ 
        printf("%d %d",t[cho],t[now]);
        bz=1; 
    } 
    return ;
}
it find(it cho,it now){
    it j=0; 
    book[cho]=1;
    if(t[cho]+t[now]==m){
        print(cho,now);
        return 0;
    }
    fr(it i=1;i<=n;i++){
        if(book[i]==0){
            book[i]=1;
            find(cho,i);
            book[i]=0;
        }
    }
    return 0;
}
it main(){
    scanf("%d",&n);
    fr(it i=1;i<=n;i++){
        book[i]=0;
        scanf("%d",&t[i]);
    }
    sort(t+1,t+1+n);
    scanf("%d",&m);
    fr(it i=1;i<=n;i++){
        find(i,0);
        if(bz==1){
            return 0;
        }
    }
    return 0;
}
4分垃圾


然而很快发现自己没加$No$,于是机智地补上了

#include <bits/stdc++.h>
#include <algorithm>
#define it int
#define fr for
#define il inline
using namespace std;
it book[100020],t[100020];
it spe[50010];
it n,m,bz=0;
it main();
it find();
it max();
it min();
il it min(it a,it b){
    if(a>b){
        return b;
    }
    else{
        return a;
    }
}
il it max(it a,it b){
    if(a>b){
        return a;
    }
    else{
        return b;
    }
}
void print(it cho,it now){
    if(bz==0){ 
        printf("%d %d",t[cho],t[now]);
        bz=1; 
    } 
    return ;
}
it find(it cho,it now){
    it j=0; 
    book[cho]=1;
    if(t[cho]+t[now]==m){
        print(cho,now);
        return 0;
    }
    fr(it i=1;i<=n;i++){
        if(book[i]==0){
            book[i]=1;
            find(cho,i);
            book[i]=0;
        }
    }
    return 0;
}
it main(){
    scanf("%d",&n);
    fr(it i=1;i<=n;i++){
        book[i]=0;
        scanf("%d",&t[i]);
    }
    sort(t+1,t+1+n);
    scanf("%d",&m);
    fr(it i=1;i<=n;i++){
        find(i,0);
        if(bz==1){
            return 0;
        }
    }
    if(bz==0){
        printf("No");
    }
    return 0;
}
6分垃圾


$mmp$怎么还是只有6分........................................................................

这时我看到了这个练习的标题

$1.11$编程基础之二分查找

mmp


 然后就出来了...

代码补充中~~~~~

原文地址:https://www.cnblogs.com/Fraction/p/8097426.html