A：签到题
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<char *,int> pii; const int N = 105; const int M = 5e4+5; const LL Mod = 199999; #define rg register #define pi acos(-1) #define INF 1e9 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } void print(int x){ if(x < 0){x = -x;putchar('-');} if(x > 9) print(x/10); putchar(x%10+'0'); } } using namespace FASTIO; void FRE(){ /*freopen("data1.in","r",stdin); freopen("data1.out","w",stdout);*/} int a[105]; int main() { int n;n = read(); for(rg int i = 1;i <= n;++i) a[i] = read(); sort(a+1,a+n+1); int len = unique(a+1,a+n+1)-a-1; printf("%d %d %d %d ",a[len]-a[len-1],a[len]-a[2],a[len-1]-a[2],a[len-1]-a[1]); // system("pause"); }
B： 暴力就行
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<char *,int> pii; const int N = 105; const int M = 5e4+5; const LL Mod = 199999; #define rg register #define pi acos(-1) #define INF 1e9 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } void print(int x){ if(x < 0){x = -x;putchar('-');} if(x > 9) print(x/10); putchar(x%10+'0'); } } using namespace FASTIO; void FRE(){ /*freopen("data1.in","r",stdin); freopen("data1.out","w",stdout);*/} int a[1005],vis[1005]; int main() { int n;n = read(); LL ans = 0; for(rg int i = 1;i <= n;++i) a[i] = read(); for(rg int i = 1;i <= n;++i) { memset(vis,0,sizeof(vis)); int ma = 0; for(rg int j = i;j <= n;++j) { if(vis[a[j]] == 0) ma++; vis[a[j]]++; ans += ma; } } printf("%lld ",ans); // system("pause"); }
C：首先，我们可以将路径上所有能到的能量都处理下来。
注意要去重，然后对这个能量做一个排序。
然后去枚举那个数作为上界，可以发现，i位置作为上限，下限选择i-x肯定差距最小（已经排过序）
然后统计最小值即可
这里用了一维数组去模拟二维，因为一开始没有看好数据范围.
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<char *,int> pii; const int N = 1e6+5; const int M = 5e4+5; const LL Mod = 199999; #define rg register #define pi acos(-1) #define INF 1e18 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } void print(int x){ if(x < 0){x = -x;putchar('-');} if(x > 9) print(x/10); putchar(x%10+'0'); } } using namespace FASTIO; void FRE(){ /*freopen("data1.in","r",stdin); freopen("data1.out","w",stdout);*/} int n,m,sx,sy,vis[N],b[4][2] = {1,0,-1,0,0,1,0,-1},top = 0; unordered_map<LL,int> mt; LL ans[N],mp[N],d,x; struct Node{int x,y,step;}; void bfs() { memset(vis,0,sizeof(vis)); queue<Node> Q; if(mp[(sx-1)*m+sy] == -1) return ; Q.push(Node{sx,sy,0}); vis[(sx-1)*m+sy] = 1; if(mp[(sx-1)*m+sy] > 0) ans[++top] = mp[(sx-1)*m+sy],mt[mp[(sx-1)*m+sy]]++; while(!Q.empty()) { Node q = Q.front(); Q.pop(); for(int i = 0;i < 4;++i) { int px = q.x+b[i][0]; int py = q.y+b[i][1]; if(px >= 1 && px <= n && py >= 1 && py <= m && q.step+1 <= d && !vis[(px-1)*m+py] && mp[(px-1)*m+py] != -1) { vis[(px-1)*m+py] = 1; Q.push(Node{px,py,q.step+1}); LL tmp = mp[(px-1)*m+py]; if(tmp != 0 && mt[tmp] == 0) ans[++top] = tmp,mt[tmp]++; } } } } int main() { int t;t = read(); while(t--) { top = 0; mt.clear(); n = read(),m = read(),sx = read(),sy = read(),d = read(),x = read(); for(rg int i = 1;i <= n;++i) for(rg int j = 1;j <= m;++j) mp[(i-1)*m+j] = read(); bfs(); if(top < x) printf("no "); else { sort(ans+1,ans+top+1); LL ma = INF; for(rg int i = x;i <= top;++i) ma = min(ma,ans[i]-ans[i-x+1]); printf("%lld ",ma); } } // system("pause"); }
D：
首先，对于一个数，如果它是最小的不能被消耗的法术。
那么，说明比他小的法术都已经被消耗了。那么对于这部分的法术，可以后缀和统计。
所以我们要从大到小排序。
那么，我们可以去枚举已经消耗了多少法术即可。
实际上，统计的是已经消耗j的法术时，最小的不能消耗的法术为a[i]时的方案数。
那么，这部分的方案数显然就是消耗掉j的方案数，那么显然这部分和前面的方案有关。
那么可以发现就是背包计数。
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<string,int> pii; const int N = 1e5+5; const int M = 2e5+5; const LL Mod = 199999; #define rg register #define pi acos(-1) #define INF 1e9 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } void print(int x){ if(x < 0){x = -x;putchar('-');} if(x > 9) print(x/10); putchar(x%10+'0'); } } using namespace FASTIO; void FRE(){/*freopen("data1.in","r",stdin); freopen("data1.out","w",stdout);*/} int a[3005]; LL dp[3005],sum[3005]; int main() { int n,m;n = read(),m = read(); for(rg int i = 1;i <= n;++i) a[i] = read(); sort(a+1,a+n+1,greater<>()); sum[n] = a[n]; for(rg int i = n-1;i >= 1;--i) sum[i] = sum[i+1]+a[i]; LL ans = 0; dp[0] = 1; for(rg int i = 1;i <= n;++i)//枚举最小的无法使用的法术 { for(rg int j = m;j >= 0;--j)//枚举剩余值j { if(m < a[i]) break;//过滤过大值 if(a[i] > m-sum[i+1]-j && j+sum[i+1] <= m) ans += dp[j];//第一个判断，当前值不够用a[i].第二个判断值不超过m } for(rg int j = m;j >= a[i];--j) dp[j] += dp[j-a[i]]; } printf("%lld ",ans); // system("pause"); }