BestCoder Round #62 (div.2) HDOJ5562 Clarke and food(脑洞)
2015-11-18 10:23
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Clarke and food
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 501 Accepted Submission(s): 292
Problem Description
Clarke is a patient with multiple personality disorder. One day, Clarke turned into a cook, was shopping for food.
Clarke has bought n food.
The volume of the ith
food is vi.
Now Clarke has a pack with volume V.
He wants to carry food as much as possible. Tell him the maxmium number he can brought with this pack.
Input
The first line contains an integer T(1≤T≤10),
the number of the test cases.
For each test case:
The first line contains two integers n,V(1≤n≤105,1≤V≤109).
The second line contains n integers,
the ith
integer denotes vi(1≤vi≤109).
Output
For each test case, print a line with an integer which denotes the answer.
Sample Input
1
3 5
1 3 4
Sample Output
2
Hint:
We can carry 1 and 3, the total volume of them is 5.
题目链接:点击打开链接
脑洞题, 本以为是背包, 读入数据后排序, 从小到大一直拿, 拿到不能拿为止可以使装食物数量最大.
AC代码:
#include "iostream" #include "cstdio" #include "cstring" #include "algorithm" #include "utility" #include "map" #include "set" #include "vector" using namespace std; typedef long long ll; const int MAXN = 1e5 + 5; int n, v, a[MAXN]; int main(int argc, char const *argv[]) { int t; scanf("%d", &t); while(t--) { scanf("%d%d", &n, &v); for(int i = 0; i < n; ++i) scanf("%d", &a[i]); sort(a, a + n); int sum = 0, ans = 0; for(int i = 0; i < n; ++i) { sum += a[i]; if(sum <= v) ans++; else break; } printf("%d\n", ans); } return 0; }
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