CodeForces 366A Dima and Guards（结构体，数学）

Description
Nothing has changed since the last round. Dima and Inna still love each other and want to be together. They've made a deal with Seryozha and now they need to make a deal with the dorm guards...

There are four guardposts in Dima's dorm. Each post contains two guards (in Russia they are usually elderly women). You can bribe a guard by a chocolate bar or a box of juice. For each guard you know the minimum price of the chocolate bar she can accept as a gift and the minimum price of the box of juice she can accept as a gift. If a chocolate bar for some guard costs less than the minimum chocolate bar price for this guard is, or if a box of juice for some guard costs less than the minimum box of juice price for this guard is, then the guard doesn't accept such a gift.

In order to pass through a guardpost, one needs to bribe both guards.

The shop has an unlimited amount of juice and chocolate of any price starting with 1. Dima wants to choose some guardpost, buy one gift for each guard from the guardpost and spend exactlyn rubles on it.

Help him choose a post through which he can safely sneak Inna or otherwise say that this is impossible. Mind you, Inna would be very sorry to hear that!

Input
The first line of the input contains integer n (1 ≤ n ≤ 105) — the money Dima wants to spend. Then follow four lines describing the guardposts. Each line contains four integers a, b, c, d (1 ≤ a, b, c, d ≤ 105) — the minimum price of the chocolate and the minimum price of the juice for the first guard and the minimum price of the chocolate and the minimum price of the juice for the second guard, correspondingly.

Output
In a single line of the output print three space-separated integers: the number of the guardpost, the cost of the first present and the cost of the second present. If there is no guardpost Dima can sneak Inna through at such conditions, print -1 in a single line.

The guardposts are numbered from 1 to 4 according to the order given in the input.

If there are multiple solutions, you can print any of them.

Sample Input
Input
10
5 6 5 6
6 6 7 7
5 8 6 6
9 9 9 9
Output
1 5 5
Input
10
6 6 6 6
7 7 7 7
4 4 4 4
8 8 8 8
Output
3 4 6
Input
5
3 3 3 3
3 3 3 3
3 3 3 3
3 3 3 3
Output
-1
Hint
Explanation of the first example.

The only way to spend 10 rubles to buy the gifts that won't be less than the minimum prices is to buy two 5 ruble chocolates to both guards from the first guardpost.

Explanation of the second example.

Dima needs 12 rubles for the first guardpost, 14 for the second one, 16 for the fourth one. So the only guardpost we can sneak through is the third one. So, Dima can buy 4 ruble chocolate for the first guard and 6 ruble juice of the second guard.\

代码：

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;struct st{int j;int k;int l;int o;int min;}a[5];int main(){  int n;  while(scanf("%d",&n)!=EOF)  {  memset(a,0,sizeof(a));  int u=0;  for(int i=1;i<=4;i++)  {  scanf("%d %d %d %d",&a[i].j,&a[i].k,&a[i].l,&a[i].o);  a[i].min+=(min(a[i].j,a[i].k)+min(a[i].l,a[i].o));  }  for(int i=1;i<=4;i++)  {  if(a[i].min<=n)  {  printf("%d %d %d\n",i,min(a[i].j,a[i].k),n-min(a[i].j,a[i].k));  u++;  break;  }  }  if(u==0)  printf("-1\n");}    return 0;}

题意：王子去找公主，但是要找到公主需要通过四扇门中的一扇，每扇门有两个守卫，他必须去用巧克力或者果汁去贿赂某个门的两个守卫才能找到公主。已知王子准备花多少钱还有每个守卫需要贿赂的巧克力或者果汁的多少。要求输出通过的是拿扇门还有给每个卫士贿赂的多少。

思路：判断每个卫士需要最小贿赂是多少，加起来后判断是否小于王子要花的钱。如果四扇门都打不开，可怜的王子就要孤独了，所以输出-1.（这里有个坑，就是王子的钱必须花完……所以都合格了就直接输出第一个卫士的和王子的钱减去卫士的）。