• HDU 2141


    题意

    给出A, B, C三个数组, 长度分别为l, m, n ( 1 <= l, m, n <= 500 )
    给出s (1 <= s <= 1000)个查询, 查找是否存在x使得在A, B, C中各取一个数使得a+b+c == x

    思路

    500*500*500*1000 复杂度太高无法暴力求解
    考虑到用二分查找
    先将前两个数列相加产生ab[]数列,这时ab+c = x,那么ab= x-c,每次查询x时,遍历c中的所有数,求得x-c,在ab数列中二分查找。
    手写二分查找或者用STL中的binary_search都可以

    二分查找变形合集 :
    你真的会写二分查找吗?

    STL中的二分查找:
    STL之二分查找(binary_search(),lower_bound(),upper_bound() )
    upper_bound 和lower_bound 的返回值问题

    AC代码(手写二分查找)

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    #define mst(a) memset(a, 0, sizeof(a))
    using namespace std;
    const int maxn = 500+5;
    typedef unsigned long long ll;
    ll a[maxn], b[maxn], c[maxn], ab[maxn*maxn];
    
    void init(){
        mst(a);mst(b);mst(c);mst(ab);
    }
    
    int binsearch(ll arr[], int len, int key)
    {
        int left = 0;
        int right = len - 1;
        int mid;
        while(left <= right){
            mid = (left + right)/2;
            if (key <= arr[mid])  right = mid - 1;
            else   left = mid + 1;
        }
        if(left < len && arr[left] == key)
            return left;
        return -1;
    }
    
    int main()
    {
        int l, m, n, k, t;
        int kase = 0;
        while( ~scanf("%d%d%d",&l, &m, &n) ){
            init();
            for( int i = 0; i < l; i++ )   scanf("%llu",&a[i]);
            for( int i = 0; i < m; i++ )   scanf("%llu",&b[i]);
            for( int i = 0; i < n; i++ )   scanf("%llu",&c[i]);
            printf("Case %d:
    ",++kase);
            sort(c, c+n);
            k = -1;
            for( int i = 0; i < l; i++ )
                for( int j = 0; j < m; j++ )
                    ab[++k] = a[i] + b[j];
            sort(ab, ab+k);
            int len = k;
            scanf("%d",&t);
            while(t--){
                scanf("%d",&k);
                bool flag = false;
                for( int i = 0; i < n; i++ ){
                    int xx = k - c[i];
                    if( binsearch( ab, len, xx ) != -1 ){
                        printf("YES
    ");
                        flag = true;
                        break;
                    }
                }
                if(!flag) printf("NO
    ");
            }
        }
        return 0;
    }
    

    AC代码(STL)

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    
    #define mst(a) memset(a, 0, sizeof(a))
    using namespace std;
    const int maxn = 500+5;
    typedef long long ll;
    ll a[maxn], b[maxn], c[maxn], ab[maxn*maxn];
    
    void init(){
        mst(a);mst(b);mst(c);mst(ab);
    }
    
    int main()
    {
        int l, m, n, k, t;
        int kase = 0;
        while( ~scanf("%d%d%d",&l, &m, &n) ){
            init();
            for( int i = 1; i <= l; i++ )   scanf("%lld",&a[i]);
            for( int i = 1; i <= m; i++ )   scanf("%lld",&b[i]);
            for( int i = 1; i <= n; i++ )   scanf("%lld",&c[i]);
            printf("Case %d:
    ",++kase);
            sort(c+1, c+n+1);
            k = 0;
            for( int i = 1; i <= l; i++ )
                for( int j = 1; j <= m; j++ )
                    ab[++k] = a[i] + b[j];
            sort(ab+1, ab+k+1);
            int len = k;
            scanf("%d",&t);
            while(t--){
                scanf("%d",&k);
                bool flag = false;
                for( int i = 1; i <= n; i++ ){
                    int xx = k - c[i];
                    if( binary_search( ab+1, ab+len+1, xx ) != 0 ){
                        printf("YES
    ");
                        flag = true;
                        break;
                    }
                }
                if(!flag) printf("NO
    ");
            }
        }
        return 0;
    }
    
  • 相关阅读:
    javascript-类型、值和变量
    词法结构
    前端js保存页面为图片下载到本地
    js生成带log的二维码(qrcodejs)
    工具函数
    css 水平垂直居中
    vue中的keep-alive
    vuex 的使用
    Spring源码解析
    Spring源码解析
  • 原文地址:https://www.cnblogs.com/JinxiSui/p/9740561.html
Copyright © 2020-2023  润新知