张小生的先生的博客

emmm，本以为学期末复习是一种不刷题的借口，最后发现只不过是自己偷懒的一种借口。～～A. Short Substrings题解：找一种合适的模拟方法做即可。int main(){ int t; cin >> t; while(t--) { string s,ans=""; cin >> s; ans+=s[0]; for(int i=1;i<s.length();i+=2) ans+=s[i]; cout << ans &lt

题目链接：D. Cut and Stick题意给你一个长度为n的序列，有q次查询，每次查询有一个区间范围[l,r]，问如何划分a[l]~a[r]，使得各个子序列每个数出现次数不超过⌈len2⌉{ \lceil \frac{len}{2} \rceil}⌈2len​⌉，len为划分后的区间长度。注意子序列不一定连续。题解首先这道题我们需要知道任意一个区间内出现次数超过⌈len2⌉{ \lceil \frac{len}{2} \rceil}⌈2len​⌉的x是谁。如果出现次数都没有超过说明没有必要划

Educational Codeforces Round 104 (Rated for Div. 2)A. Arenaint a[maxn];int main(){ ios; int t; cin >> t; while (t--) { int n; cin >> n; int minn=inf; for(int i=1;i<=n;i++) { cin >> a[i]; minn=min(minn,a[i]);} int ans=

Codeforces Round #704 (Div. 2)A. Three swimmersint main(){ ios; int t; cin >> t; while (t--) { ll p,a,b,c; cin >> p >> a >> b >> c; ll t1=p/a+(p%a==0?0:1); ll t2=p/b+(p%b==0?0:1); ll t3=p/c+(p%c==0?0:1); t1

Codeforces Round #702 (Div. 3)A. Dense Arrayint a[maxn];int main(){ ios; int t; cin >> t; while (t--) { int n; cin >> n; for(int i=1;i<=n;i++) cin >> a[i]; int pre=a[1]; int ans=0; for(int i=2;i<=n;i++) { in

Educational Codeforces Round 105 (Rated for Div. 2)A. ABC String枚举ABC分别为“（ ”和 “ ）”的情况，然后判断是否为合法字符串。map<char,int> book;string tt="()";string s; bool judge(string ss){ int num=0; for(int i=0;i<ss.length();i++) { if(ss[i]=

Codeforces Round #698 (Div. 2)A. Nezzar and Colorful Balls找数字出现次数最多的个数map<int,int> book;int a[maxn];int main(){ int t; cin >> t; while (t--) { book.clear(); int n; cin >> n; for(int i=1;i<=n;i++) { cin >> a[

Codeforces Round #693 (Div. 3)A. Cards for Friends按照题目递归模拟即可int f(int a,int b){ if(a%2 && b%2) return 1; if(a%2==0) return 2*f(a/2,b); else return 2*f(a,b/2);}int main(){ int t; cin >> t; while(t--) { int w,h,n; cin >&gt

题目链接：Codeforces Round #687 (Div. 2)A. Prison Break找四个角到(r,c)的最大距离即可int main(){ int t; cin >> t; while(t--) { int n,m,r,c; cin >> n >> m >> r >> c; int ans=0; ans=max(ans,r-1+c-1); ans=max(ans,r-1+m-c); ans=

题目链接：Educational Codeforces Round 99 (Rated for Div. 2)A. Strange Functions转换题意就是判断字符串长度int main(){ int n; cin >> n; while(n--) { string s; cin >> s; cout << s.length() << endl; }}B. Jumpsint main(){ int t; sc

题目链接：Codeforces Round #686 (Div. 3)A. Special Permutation整体向前循环移一位int main(){ int t; cin >> t; while(t--) { int n; cin >> n; for(int i=1;i<=n;i++) { if((i+1)%(n+1)==0) cout << 1 << ' '; else cout <&lt

题目链接：Codeforces Round #685 (Div. 2)A. Subtract or Divide贪心，不难发现除了2的偶数仅有两次操作就行，除了1和3的奇数仅有三次操作；剩下的特判即可。int main(){ int t; cin >> t; while(t--) { int n; cin >> n; if(n%2==0) { if(n==2) cout << 1 << endl; else cout

题目链接：Codeforces Round #684 (Div. 2)A. Buy the String贪心 int main(){ int t; cin >> t; while(t--) { int n,c0,c1,h; cin >> n >> c0 >> c1 >> h; string s; cin >> s; int ans=0; if(c0>c1) { if(h+c1<

题目链接：Educational Codeforces Round 98 (Rated for Div. 2)A. Robot Program贪心int main(){ int t; cin >> t; while(t--) { int x,y; cin >> x >> y; if(fabs(x-y)<=1) { cout << x+y << endl; continue; } else

A. XORwice题意输入a和b，求 (????⊕????) + (????⊕????) (x为任意值) 的最小值。题解要使(????⊕????) + (????⊕????)最小，言外之意就是将a和b转化为二进制下的公共的1部分被异或，答案其实就是a^b。int main(){ int t; cin >> t; while(t--) { int a,b; cin >> a >> b; cout << (a^b) <&lt