APAC 2013 部分题解 | 训练记录
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目录
A . The Alphabet Sticker
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
int main()
int _;cin>>_;
while(_--)
string s;cin>>s;
ll l=0,r=s.size()-1;
while(s[l]=='?') l++;
while(s[r]=='?') r--;
ll ans=1ll,tmp=0;
for(int i=l;i<=r;i++)
if(s[i]=='?')
tmp=i;
while(s[tmp]=='?') tmp++;
ll now=tmp-i+1;
if(s[i-1]!=s[tmp]) ans=ans*now%mod;
i=tmp;
cout<<ans<<endl;
return 0;
C . Increasing Shortest Path
排序升序加边跑dp
/*** keep hungry and calm PushyTao!***/
struct node
int u, v, w;
friend bool operator<(node a, node b) return a.w < b.w;
e[3007];
int dp[157][157][157];
int mp[157][157];
int main()
int _ = read;
while (_--)
int n = read, m = read, q = read;
memset(mp, 0x3f3f3f3f, sizeof mp);
for (int i = 1; i <= m; i++)
// int u = read, v = read, w = read;
// mp[u][v] = min(mp[u][v], w);
e[i].u = read, e[i].v = read, e[i].w = read;
/**
int idx = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if(mp[i][j] < 0x3f3f3f3f)
idx ++;
e[idx].u = i;
e[idx].v = j;
e[idx].w = mp[i][j];
**/
sort(e + 1, e + 1 + m);
memset(dp, 0x3f3f3f3f, sizeof dp);
for (int i = 1; i <= n; i++) dp[i][i][0] = 0; //自己到自己的距离是0
for (int i = 1; i <= m; i++)
for (int qi = 1; qi <= n; qi++)
for (int skp = 1; skp <= n - 1; skp++) // len max n-1
dp[qi][e[i].v][skp] = min(dp[qi][e[i].v][skp], dp[qi][e[i].u][skp - 1] + e[i].w);
while (q--)
int x = read, y = read, c = read;
c = min(c, n - 1);
itn ans = 0x3f3f3f3f;
for (int i = 0; i <= c; i++) ans = min(ans, dp[x][y][i]);
if (ans == 0x3f3f3f3f)
puts("-1");
else
printf("%d\\n", ans);
return 0;
/**
1
8 9 3
1 2 1
2 3 2
3 4 3
4 5 12
5 8 7
1 6 8
6 4 9
1 7 5
7 4 4
1 4 2
1 4 3
1 4 1
**/
D . Cup of Cowards
#include <bits/stdc++.h>
typedef long long ll;
#pragma region Debug
template<typename T>
std::istream& operator>> (std::istream& in, std::vector<T>& vt)
for (auto& i : vt) in >> i;
return in;
template<typename T>
std::ostream& operator<< (std::ostream& out, const std::vector<T>& vt)
for (auto& i : vt) out << i << " ";
out << std::endl;
return out;
void printAll()
template<typename T1, typename... T2>
void printAll(const T1& first, const T2&... second)
std::cout << first;
if (sizeof...(second))
std::cout << ", ";
printAll(second...);
template<typename T1, typename... T2>
void print(const T1& first, const T2&... second)
std::cout << "[";
printAll(first, second...);
std::cout << "]" << std::endl;
#pragma endregion
#define N (int)(5e2 + 10)
struct Node
int h;
long long d, c;
friend bool operator< (Node& A, Node& B)
if (A.d * B.c != B.d * A.c)
return A.d * B.c > B.d * A.c;
return A.c < B.c;
;
long long L;
long long sufDam[N];
std::vector<Node> v(5);
long long cost, dam;
void dfs(int dep, ll curCost, ll curDam)
if (curDam >= L)
if (curCost < cost)
cost = curCost;
dam = curDam;
else if (curCost == cost)
dam = std::min(dam, curDam);
return;
if (dep == 5) return;
if (curDam + sufDam[dep] < L) return;
ll rouDam = L - curDam;
ll cnt = (rouDam + (v[dep].d - 1)) / v[dep].d;
if ((cnt - 1) * v[dep].c + curCost > cost) return;
if (cnt > v[dep].h) cnt = v[dep].h;
for (int j = cnt; j >= 0; j--)
dfs(dep + 1, curCost + 1LL * j * v[dep].c, curDam + 1LL * j * v[dep].d);
int main()
int T;
std::cin >> T;
while (T--)
std::cin >> L;
for (auto& [h, d, c] : v)
std::cin >> h >> d >> c;
sort(v.begin(), v.end());
sufDam[4] = 1LL * v[4].h * v[4].d;
for (int i = 3; i >= 0; i--)
sufDam[i] = sufDam[i + 1] + 1LL * v[i].h * v[i].d;
cost = LLONG_MAX, dam = LLONG_MAX;
dfs(0, 0, 0);
if (cost == LLONG_MAX)
std::cout << "We are doomed!!" << std::endl;
else
std::cout << cost << " " << dam << std::endl;
return 0;
/*
2 3 4
3 1 2
2 3 2
31 15
D题题意：
输入
T
L
五组H, D, C
L表示怪物的总血量
五个玩家
H表示当前玩家可以攻击的次数限制
D表示每次攻击造成的伤害
C表示花费
问杀死怪物的最小花费和最小伤害
*/
E . Balloons Colors
int a[107];
int main()
int _ = read;
while(_--)
int n = read;
a[1] = read,a[n] = read;
int f1 = 0, f2 = 0;
for(int i=1;i<=n;i++)
int x = read;
if(x == a[1] && i == 1) f1 = 1;
if(x == a[n] && i == n) f2 = 1;
if(f1 && f2) puts("BOTH");
else if(f1) puts("EASY");
else if(f2) puts("HARD");
else puts("OKAY");
return 0;
F . NASSA’s Robot
int main()
int _ = read;
while(_--)
string s;
cin >> s;
int _not = 0,x = 0,y = 0;
for(int i=0; i<s.size(); i++)
if(s[i] == 'U') y ++;
else if(s[i] == 'D') y --;
else if(s[i] == 'L') x --;
else if(s[i] == 'R') x ++;
else _not ++;
printf("%d %d %d %d\\n",x-_not,y-_not,x+_not,y+_not);
return 0;
G . The Stones Game
int a[107];
int main()
int _ = read;
while(_--)
int n = read, m = read, x = read;
n -= x;
if(n % m) puts("NO");
else puts("YES");
return 0;
I . Omar Loves Candies
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
int a[1100][1100],sum[1100][1100];
inline int Max(int a, int b)
return a > b ? a : b;
int main()
int _;scanf("%d",&_);
while(_--)
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
scanf("%d",&a[i][j]);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
sum[i][j]=a[i][j]+sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
int ans=-1e9;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
ans=Max(ans,sum[n][m]-sum[n][j-1]-sum[i-1][m]+sum[i-1][j-1]);
//cout<<i<<" "<<j<<" "<<ans<<endl;
printf("%d\\n",ans);
return 0;
J . Modified LCS
#include <bits/stdc++.h>
typedef long long ll;
#pragma region Debug
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