Codeforces Round #381 (Div. 2) A B C
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A:
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for arubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
1 1 3 4
3
6 2 1 1
1
4 4 4 4
0
999999999 1000000000 1000000000 1000000000
1000000000
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
这题其实可以直接上暴力的。。。还去分类讨论了。。而且速度还那么慢。。。而且居然还漏了一种情况
智障啊。。
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <map>#include <set>#include <cstdlib>#include <queue>#include <stack>using namespace std;int main(){ long long n,a,b,c,now; scanf("%I64d%I64d%I64d%I64d",&n,&a,&b,&c); if(n==0){ now = min(4*a,2*b); now = min(now,4*c); now = min(now,2*a+b); now = min(now,a+c); now = min(now,b+2*c); printf("%I64d\n",now); return 0; } if(n%4==1){ now = min(a*3,c); now = min(now,a+b); printf("%I64d\n",now); }else if(n%4==2){ now = min(a*2,b); now = min(now,2*c); printf("%I64d\n",now); }else if(n%4==3){ now = min(a,3*c); now = min(now,b+c); printf("%I64d\n",now); }else{ printf("0\n"); } return 0;}
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <map>#include <set>#include <cstdlib>#include <queue>#include <stack>using namespace std;const int maxn = 110;int a[maxn];int main(){ int n,m,i,j,l,r; scanf("%d%d",&n,&m); for(i=1;i<=n;i++){ scanf("%d",&a[i]); } int ans = 0,now = 0; for(i=1;i<=m;i++){ scanf("%d%d",&l,&r); now = 0; for(j=l;j<=r;j++){ now += a[j]; } if(now>0){ ans += now; } } printf("%d\n",ans); return 0;}
鸟神给了提示才做的。。。。自己好智障啊。。。
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <map>#include <set>#include <cstdlib>#include <queue>#include <stack>using namespace std;const int maxn = 1e5;int res[maxn];int main(){ int n,m,i,j,l,r; scanf("%d%d",&n,&m); int minn = 1000100; int l0,r0; for(i=1;i<=m;i++){ scanf("%d%d",&l,&r); if(r-l+1<minn){ minn = r-l+1; l0 = l; r0 = r; } } printf("%d\n",minn); int len = minn-1; for(i=l0;i<=r0;i++){ res[i] = len--; } len = 0; for(i=l0-1;i>=1;i--){ res[i] = len; len = (len+1+minn)%minn; } len = minn-1; for(i=r0+1;i<=n;i++){ res[i] = len; len = (len-1+minn)%minn; } for(i=1;i<=n;i++){ printf("%d ",res[i]); } return 0;}
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