SDUT ACM 2157 Greatest Number
Greatest Number
Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^
题目描述
Saya likes math, because she think math can make her cleverer.
One day, Kudo invited a very simple game:
Given N integers, then the players choose no more than four integers from them (can be repeated) and add them together. Finally, the one whose sum is the largest wins the game. It seems very simple, but there is one more condition: the sum shouldn’t larger than a number M.
Saya is very interest in this game. She says that since the number of integers is finite, we can enumerate all the selecting and find the largest sum. Saya calls the largest sum Greatest Number (GN). After reflecting for a while, Saya declares that she found the GN and shows her answer.
Kudo wants to know whether Saya’s answer is the best, so she comes to you for help.
Can you help her to compute the GN?
One day, Kudo invited a very simple game:
Given N integers, then the players choose no more than four integers from them (can be repeated) and add them together. Finally, the one whose sum is the largest wins the game. It seems very simple, but there is one more condition: the sum shouldn’t larger than a number M.
Saya is very interest in this game. She says that since the number of integers is finite, we can enumerate all the selecting and find the largest sum. Saya calls the largest sum Greatest Number (GN). After reflecting for a while, Saya declares that she found the GN and shows her answer.
Kudo wants to know whether Saya’s answer is the best, so she comes to you for help.
Can you help her to compute the GN?
输入
The input consists of several test cases.
The first line of input in each test case contains two integers N (0<N≤1000) and M(0<M≤ 1000000000), which represent the number of integers and the upper bound.
Each of the next N lines contains the integers. (Not larger than 1000000000)
The last case is followed by a line containing two zeros.
The first line of input in each test case contains two integers N (0<N≤1000) and M(0<M≤ 1000000000), which represent the number of integers and the upper bound.
Each of the next N lines contains the integers. (Not larger than 1000000000)
The last case is followed by a line containing two zeros.
输出
For each case, print the case number (1, 2 …) and the GN.
Your output format should imitate the sample output. Print a blank line after each test case.
Your output format should imitate the sample output. Print a blank line after each test case.
示例输入
2 10 100 2 0 0
示例输出
Case 1: 8
//开始用搜索或者顺序查找，都超时，改用二分查找，110ms过了
#include <stdio.h>
#include <algorithm>
int n, m, item = 1;
int q[1000010], a[1010];
int main()
{
a[0] = 0;
while(scanf("%d %d", &n, &m) != EOF && (n || m))
{
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
//因为a[0] == 0，所以a[i] + a[j]包括了0个数、1个数、2个数的和
std::sort(a, a+n+1);
int rear = 0;
for(int i = 0; i <= n; i++)
for(int j = i; j <= n; j++)//注意j = i；避免重复计算
if(a[i] + a[j] <= m)
q[rear++] = a[i] + a[j];
std::sort(q, q+rear);
int ans = 0;
for(int i = 0; i < rear; i++)
{
int left = i, right = rear-1; //同样left = i，避免重复计算
while(left < right)
{
int mid = (left + right) / 2;
if(q[i] + q[mid] > m)
right = mid - 1;
else left = mid + 1;
}
if(q[i] + q[left] > ans && q[i] + q[left] <= m) //这里要再判断一下是否小于m
ans = q[i] + q[left];
}
printf("Case %d: %d\n\n", item++, ans);
}
return 0;
}
