codeforces 415B. Mashmokh and Tokens
http://codeforces.com/problemset/problem/415/B
Bimokh is Mashmokh's boss. For the following n days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives backw tokens then he'll get
dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He hasn numbers x1, x2, ..., xn. Numberxi is the number of tokens given to each worker on thei-th day. Help him calculate for each of n
Input
The first line of input contains three space-separated integers n, a, b (1 ≤ n ≤ 105; 1 ≤ a, b ≤ 109). The second line of input containsn space-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 109).
Output
Output n space-separated integers. The i-th of them is the number of tokens Mashmokh can save on the i-th day.
Sample test(s)
Input
5 1 412 6 11 9 1
Output
0 2 3 1 1
Input
3 1 21 2 3
Output
1 0 1
Input
1 1 11
Output
0
什么都不用说，唯一可能需要的是有道词典。
#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
const int N=1e5+10;
LL n,a,b;
LL w[N];
int main()
{
while(cin>>n>>a>>b){
for(int i=0;i<n;i++) scanf("%I64d",&w[i]);
for(int i=0;i<n;i++){
LL t1=w[i]*a/b;
LL t2=t1*b;
LL ans=t2/a;
while(ans*a<t2){
ans++;
}
if(i!=n-1) printf("%I64d ",w[i]-ans);
else printf("%I64d\n",w[i]-ans);
}
}
return 0;
}
codeforces 413 C. Jeopardy!
http://codeforces.com/contest/413/problem/C
Jeopardy!' is an intellectual game where players answer questions and earn points. Company Q conducts a simplified 'Jeopardy!' tournament among the best IT companies. By a lucky coincidence, the old rivals made it to the finals: company R1 and company R2.
The finals will have n questions, m of them are auction questions and n - m of them are regular questions. Each question has a price. The price of thei-th question is ai
The game will go as follows. First, the R2 company selects a question, then the questions are chosen by the one who answered the previous question correctly. If no one answered the question, then the person who chose last chooses again.
All R2 employees support their team. They want to calculate what maximum possible number of points the R2 team can get if luck is on their side during the whole game (they will always be the first to correctly answer questions). Perhaps you are not going to be surprised, but this problem was again entrusted for you to solve.
Input
The first line contains two space-separated integers n andm (1 ≤ n, m ≤ 100; m ≤ min(n, 30)) — the total number of questions and the number of auction questions, correspondingly. The second line containsn space-separated integers a1, a2, ..., an(1 ≤ ai ≤ 107) — the prices of the questions. The third line containsm distinct integers bi (1 ≤ bi ≤ n) — the numbers of auction questions. Assume that the questions are numbered from1 to n.
Output
In the single line, print the answer to the problem — the maximum points the R2 company can get if it plays optimally well. It is guaranteed that the answer fits into the integer 64-bit signed type.
Sample test(s)
Input
4 11 3 7 5 3
Output
18
Input
3 210 3 8 2 3
Output
40
Input
2 2100 200 1 2
Output
400
有一个特点，按价值排好序后的竞争问题，如果不改变最后（最大价值）的问题prize，能答对的话，新得到的prize和一定大于前面所有的竞争问题的prize。依据它做出比较。
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N=110;
typedef long long LL;
LL a[N],que[35];
int top,b;
int main()
{
//freopen("cin.txt","r",stdin);
int n,m;
while(cin>>n>>m){
top=0;
for(int i=0;i<n;i++){
scanf("%I64d",&a[i]);
}
for(int i=0;i<m;i++){
scanf("%d",&b);
que[top++]=a[b-1];
a[b-1]=0;
}
sort(que,que+m);
int bot=0;
LL sum =0;
for(int i=0;i<n;i++){
if(a[i]) sum+=a[i];
}
LL temp=0,maxm=-1;
for(int i=top-1;i>=0;i--){
temp=sum;
for(int j=top-1;j>=i;j--){
temp=temp+que[j];
}
temp<<=i;
if(maxm<temp){
maxm=temp;
}
}
temp=sum;
for(int i=0;i<top;i++){
if(temp>que[i]){
temp<<=1;
}
else temp+=que[i];
}
maxm=max(maxm,temp);
printf("%I64d\n",maxm);
}
return 0;
}
codeforces 416 C. Booking System
http://codeforces.com/problemset/problem/416/C(因为粘贴过来出现了各种格式错误，我就不贴了）
经典的贪心，按照钱的多少总排序，然后在钱相等的情况下再按人数从小到大排序，最后对tables人数从小到大排序。即资源消耗最小化，收益最大化
。（它和01背包不同，因为桌子和团队的关系是1对1的）
#include <iostream>//贪心和暴力相比更好写且思路清晰
#include <cstdio>
#include <algorithm>
using namespace std;
int n,m,all;
struct node{
int order,peo,money;
bool vis;
}f[1010];
int cmp(node a,node b){
if(a.money!=b.money){
return a.money>b.money;
}
return a.peo<b.peo; //贪心要贪的彻底！
}
struct node1{
int order,peo;
}t[1010];
int cmp1(node1 a,node1 b){
return a.peo<b.peo;
}
int ans[1010][2]; //quest number and table number
int main()
{
//freopen("cin.txt","r",stdin);
while(cin>>n){
for(int i=0;i<n;i++){
scanf("%d%d",&f[i].peo,&f[i].money);
f[i].order=i+1;
f[i].vis=false;
}
sort(f,f+n,cmp);
scanf("%d",&m);
for(int i=0;i<m;i++){
scanf("%d",&t[i].peo);
t[i].order=i+1;
}
sort(t,t+m,cmp1);
int top=0,sum=0;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(f[j].peo<=t[i].peo&&f[j].vis==false){
ans[top][0]=f[j].order;
ans[top++][1]=t[i].order;
f[j].vis=1;
sum+=f[j].money;
break;
}
}
}
printf("%d %d\n",top,sum);
for(int i=0;i<top;i++){
printf("%d %d\n",ans[i][0],ans[i][1]);
}
}
return 0;
}
codeforces 416 D. Population Size
http://codeforces.com/problemset/problem/416/D
由前两个非-1的数字可以决定一段等差数列，所以可以直接遍历数组，看是否能够合并到前面的等差数列中。然而我一直卡死在79
个样例中了 T_T 自己写的代码有问题。。。
