time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Nikolay has a lemons, b apples and c pears. He decided to cook a compote. According to the recipe the fruits should be in the ratio 1:?2:?4. It means that for each lemon in the compote should be exactly 2 apples and exactly 4 pears. You can’t crumble up, break up or cut these fruits into pieces. These fruits — lemons, apples and pears — should be put in the compote as whole fruits.

Your task is to determine the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. It is possible that Nikolay can’t use any fruits, in this case print 0.

Input
The first line contains the positive integer a (1?≤?a?≤?1000) — the number of lemons Nikolay has.

The second line contains the positive integer b (1?≤?b?≤?1000) — the number of apples Nikolay has.

The third line contains the positive integer c (1?≤?c?≤?1000) — the number of pears Nikolay has.

Output
Print the maximum total number of lemons, apples and pears from which Nikolay can cook the compote.

Examples
input
2
5
7
output
7
input
4
7
13
output
21
input
2
3
2
output
0
Note
In the first example Nikolay can use 1 lemon, 2 apples and 4 pears, so the answer is 1?+?2?+?4?=?7.

In the second example Nikolay can use 3 lemons, 6 apples and 12 pears, so the answer is 3?+?6?+?12?=?21.

In the third example Nikolay don’t have enough pears to cook any compote, so the answer is 0.

【題目鏈接】:http://codeforces.com/contest/746/problem/A

【題解】

a+1,b+2,c+4;
看看能不能加就好;
能加一直加;
最后全部加起來就是答案;

【完整代碼】

#include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long #define rep1(i,a,b) for (int i = a;i <= b;i++) #define rep2(i,a,b) for (int i = a;i >= b;i--) #define mp make_pair #define pb push_back #define fi first #define se second #define rei(x) scanf("%d",&x) #define rel(x) scanf("%I64d",&x)typedef pair<int,int> pii; typedef pair<LL,LL> pll;//const int MAXN = x; const int dx[9] = {0,1,-1,0,0,-1,-1,1,1}; const int dy[9] = {0,0,0,-1,1,-1,1,-1,1}; const double pi = acos(-1.0);int a[4]; int ma,mb,mc;int main() {//freopen("F:\\rush.txt","r",stdin);rei(ma);rei(mb);rei(mc);a[1] = 1,a[2] = 2,a[3] = 4;LL x = 0,y = 0,z = 0;while (x+a[1] <= ma && y+a[2] <= mb && z+a[3] <= mc){x+=a[1],y+=a[2],z+=a[3];}cout << x + y +z;return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/AWCXV/p/7626801.html

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