codeforces 439A Devu, the Singer and Churu, the Joker
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Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing n songs, ith song will take ti minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than d minutes;
- Devu must complete all his songs;
- With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
The first line contains two space separated integers n, d (1 ≤ n ≤ 100; 1 ≤ d ≤ 10000). The second line contains n space-separated integers: t1, t2, ..., tn (1 ≤ ti ≤ 100).
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
3 302 2 1
5
3 202 1 1
-1
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes.
- Then Devu performs the first song for 2 minutes.
- Then Churu cracks 2 jokes in 10 minutes.
- Now Devu performs second song for 2 minutes.
- Then Churu cracks 2 jokes in 10 minutes.
- Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
标签虽然要用贪心,但是不用也可以!
#include<stdio.h>int main(){ int n,d,s=0,x; scanf("%d%d",&n,&d); for(int i=0; i<n; i++) { scanf("%d",&x); s+=x; } int w=(n-1)*10; if(w+s>d) { printf("-1\n"); return 0; } if(d-w-s>=5) printf("%d\n",(n-1)*2+(d-w-s)/5); else printf("%d\n",2*(n-1)); return 0;}
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