Codeforces Round #750 (Div. 2)

):稍微补一下.

A. Luntik and Concerts

void solve(){int a,b,c;scanf("%d%d%d",&a,&b,&c);printf("%d\n",(a+c)%2);
}

B. Luntik and Subsequences

void solve(){int n;scanf("%d",&n);int c1 = 0, c2 = 0;for(int i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]==1)  c1++;else if(a[i]==0)   c2++;}ll ans = 0;ans += ((ll)1<<c2)*c1;printf("%lld\n",ans);
}

C. Grandma Capa Knits a Scarf

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>using namespace std;void solve(){int n;scanf("%d",&n);string s;cin >> s;int ans = n+1;for(int i=0;i<26;i++){char c = (char)('a'+i);int cnt = 0,l = 0, r = n-1;while(l<r){if(s[l]==s[r])  l++,r--;else if(s[l]==c)    l++,cnt++;else if(s[r]==c)    r--,cnt++;else{cnt = n+1;break;}}ans = min(ans,cnt);}if(ans == n+1) puts("-1");else   printf("%d\n",ans);
}int main()
{int t;scanf("%d",&t);while(t--){solve();}return 0;
}

D. Vupsen, Pupsen and 0

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>using namespace std;void solve(){int n;scanf("%d",&n);
//  for(int i=1;i<=n;i++)    scanf("%d",&a[i]);if(n&1){int x,y,z;scanf("%d%d%d",&x,&y,&z);if((x+y)!=0) printf("%d %d %d ",-z,-z,x+y);else if((y+z)!=0)    printf("%d %d %d ",y+z,-x,-x);else printf("%d %d %d ",-y,x+z,-y);n-=3;}for(int i=1;i<=n;i+=2){int x,y;scanf("%d%d",&x,&y);printf("%d %d ",-y,x);}cout << endl;
}int main()
{int t;scanf("%d",&t);while(t--){solve();}return 0;
}

F1. Korney Korneevich and XOR (easy version)

F1,F2的差距是数据范围不一样，均为动态规划类型题.
F1可以定义dp[i]表示异或值为i的最小末尾值.

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <set>using namespace std;int f[514];
//int a[100010];
//定义dp[i]表示异或值为 i 的最小末尾值.int main()
{int n;scanf("%d",&n);//  for(int i=1;i<=n;i++)    scanf("%d",&a[i]);memset(f,0x3f,sizeof f);f[0] = 0;int x;for(int i=1;i<=n;i++){scanf("%d",&x);f[x] = min(f[x],x);for(int j=0;j<=512;j++){if(f[j]<x){f[j^x] = min(f[j^x],x);}}}set<int> st;for(int i=0;i<=512;i++){if(f[i]!=0x3f3f3f3f)    st.insert(i);}printf("%d\n",st.size());set<int>::iterator it;for(it = st.begin();it!=st.end();it++){cout << *it << " ";}cout << endl;return 0;
}

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Codeforces Round #750 (Div. 2)

):稍微补一下.

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