Google Code Jam round 1C总结

真的是很遗憾，本来47分是300名左右的，可以进入下一轮，但是自己忘记了把int该 成long long类型导致计算越界，出现了负数。我居然都没有检查输出文件，就直接交上去了，混蛋啊。。。。。

第一题：

注意不要重复计数就可以了。

#include <vector>
#include <list>
#include <limits.h>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string.h>
#include <stdlib.h>
using namespace std;

bool check(char a){
if(a=='a' || a=='e' || a=='i' || a=='o' || a=='u')
return false;
return true;
}

int main(){
freopen("A.in", "r", stdin);
freopen("1.out", "w", stdout);
int t, n;
vector<int> start;
string name;
cin>>t;
for(int j=1; j<=t; j++){
start.clear();
long long res = 0;
cin>>name>>n;
int counter = 0;
int len = name.size();
for(int i=0; i<len; i++){
if(check(name[i])==false)
counter = 0;
else
counter++;
if(counter>=n)
start.push_back(i);
}
for(int i=0; i<start.size(); i++){
long long left, right;
if(i==0)
left = start[i]-n+1+1;
else
left = start[i]-start[i-1];
if(i==start.size()-1)
right = len - start[i];
else
right = len-start[i];
//cout<<left<<" "<<right<<endl;
res += left*right;
}
cout<<"Case #"<<j<<": "<<res<<endl;
}

return 0;
}

第二题small input：

采用构造的方法，要向东走1单位，只需要想跳到西边去，紧接着跳到东边去，向北走的情形也是类似的。

#include <vector>
#include <list>
#include <limits.h>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string.h>
#include <stdlib.h>
using namespace std;

int main(){
freopen("1.in", "r", stdin);
freopen("1.out", "w", stdout);
int t, x, y;
vector<char> res;
cin>>t;
for(int i=1; i<=t; i++){
cin>>x>>y;
res.clear();
if(x>0){
for(int j=0; j<x; j++){
res.push_back('W');
res.push_back('E');
}
}
else{
for(int j=0; j<-x; j++){
res.push_back('E');
res.push_back('W');
}
}
if(y>0){
for(int j=0; j<y; j++){
res.push_back('S');
res.push_back('N');
}
}
else{
for(int j=0; j<-y; j++){
res.push_back('N');
res.push_back('S');
}
}
cout<<"Case #"<<i<<": ";
for(int j=0; j<res.size(); j++)
cout<<res[j];
cout<<endl;
}
return 0;
}

第三题small input：

一天一天的模拟就行了，注意数组下表可能取负数的情形。

#include <vector>
#include <list>
#include <limits.h>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string.h>
#include <stdlib.h>
using namespace std;

struct Node{
int di, ni, wi, ei;
int si, delta_di, delta_pi, delta_si;
};
Node node[20];

int main(){
//freopen("1.in", "r", stdin);
//freopen("1.out", "w", stdout);
int t, n;
int wall[1000005];
int middle[1000005];
cin>>t;
for(int j=1; j<=t; j++){
int res = 0;
int days = 0;
int offset = 500000;
memset(wall, 0, sizeof(wall));
memset(middle, 0, sizeof(middle));
cin>>n;
for(int i=0; i<n; i++){
cin>>node[i].di>>node[i].ni>>node[i].wi>>node[i].ei
>>node[i].si>>node[i].delta_di>>node[i].delta_pi>>node[i].delta_si;
days = max(days, node[i].di+node[i].ni*node[i].delta_di);
}
vector<pair<int, int> > v;
vector<int> height;
for(int i=0; i<=days; i++){
v.clear();
height.clear();
for(int cur=0; cur<n; cur++){
if(node[cur].ni<=0 || node[cur].di>i)
continue;
if((i-node[cur].di)%(node[cur].delta_di)!=0)
continue;
int start = node[cur].wi;
int end = node[cur].ei;
bool okay = false;
for(int k=start; k<=end; k++){
if(wall[offset+k]<node[cur].si){
okay=true;
v.push_back(make_pair(start, end));
//cout<<"start: "<<start<<" end: "<<end<<endl;
height.push_back(node[cur].si);
break;
}
if(k!=end && middle[offset+k]<node[cur].si){
okay = true;
v.push_back(make_pair(start, end));
//cout<<"start: "<<start<<" end: "<<end<<endl;
height.push_back(node[cur].si);
break;
}
}
node[cur].ni--;
node[cur].si += node[cur].delta_si;
node[cur].wi += node[cur].delta_pi;
node[cur].ei += node[cur].delta_pi;
if(okay==true)
res++;
}
int len = v.size();
for(int k=0; k<len; k++){
int start = v[k].first;
int end = v[k].second;
for(int ii=start; ii<=end; ii++){
wall[ii+offset] = max(wall[ii+offset], height[k]);
if(ii != end)
middle[ii+offset] = max(middle[ii+offset], height[k]);
}
}
}
cout<<"Case #"<<j<<": "<<res<<endl;
}
return 0;
}