8 条题解
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4
A+B Problem 那么难的题目，我们怎么可能会做呢？只能用 Python 逝世了。
a,b=map(int,input().split()) print(a+b)-
CoderCharlie
@ 2023-5-28 13:48:29
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4
//一颗资瓷区间加、区间翻转、区间求和的Splay #include <bits/stdc++.h> #define ll long long #define N 100000 using namespace std; int sz[N], rev[N], tag[N], sum[N], ch[N][2], fa[N], val[N]; int n, m, rt, x; void push_up(int x){ sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1; sum[x] = sum[ch[x][1]] + sum[ch[x][0]] + val[x]; } void push_down(int x){ if(rev[x]){ swap(ch[x][0], ch[x][1]); if(ch[x][1]) rev[ch[x][1]] ^= 1; if(ch[x][0]) rev[ch[x][0]] ^= 1; rev[x] = 0; } if(tag[x]){ if(ch[x][1]) tag[ch[x][1]] += tag[x], sum[ch[x][1]] += tag[x]; if(ch[x][0]) tag[ch[x][0]] += tag[x], sum[ch[x][0]] += tag[x]; tag[x] = 0; } } void rotate(int x, int &k){ int y = fa[x], z = fa[fa[x]]; int kind = ch[y][1] == x; if(y == k) k = x; else ch[z][ch[z][1]==y] = x; fa[x] = z; fa[y] = x; fa[ch[x][!kind]] = y; ch[y][kind] = ch[x][!kind]; ch[x][!kind] = y; push_up(y); push_up(x); } void splay(int x, int &k){ while(x != k){ int y = fa[x], z = fa[fa[x]]; if(y != k) if(ch[y][1] == x ^ ch[z][1] == y) rotate(x, k); else rotate(y, k); rotate(x, k); } } int kth(int x, int k){ push_down(x); int r = sz[ch[x][0]]+1; if(k == r) return x; if(k < r) return kth(ch[x][0], k); else return kth(ch[x][1], k-r); } void split(int l, int r){ int x = kth(rt, l), y = kth(rt, r+2); splay(x, rt); splay(y, ch[rt][1]); } void rever(int l, int r){ split(l, r); rev[ch[ch[rt][1]][0]] ^= 1; } void add(int l, int r, int v){ split(l, r); tag[ch[ch[rt][1]][0]] += v; val[ch[ch[rt][1]][0]] += v; push_up(ch[ch[rt][1]][0]); } int build(int l, int r, int f){ if(l > r) return 0; if(l == r){ fa[l] = f; sz[l] = 1; return l; } int mid = l + r >> 1; ch[mid][0] = build(l, mid-1, mid); ch[mid][1] = build(mid+1, r, mid); fa[mid] = f; push_up(mid); return mid; } int asksum(int l, int r){ split(l, r); return sum[ch[ch[rt][1]][0]]; } int main(){ //总共两个数 n = 2; rt = build(1, n+2, 0);//建树 for(int i = 1; i <= n; i++){ scanf("%d", &x); add(i, i, x);//区间加 } rever(1, n);//区间翻转 printf("%d\n", asksum(1, n));//区间求和 return 0; }-
CoderCharlie
@ 2023-5-28 13:48:08
orz
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2
GreenPayne
LV 6
@ 2023-5-13 13:24:40
区间DP
#include<bits/stdc++.h> using namespace std; int f[305][305],a[305],sum[305]; int main(){ int n; n=2;//两堆果子 for(int i=1;i<=n;i++){ scanf("%d",&a[i]); sum[i]+=sum[i-1]+a[i];//前缀和加快速度 } for(int l=1;l<=n;l++){//注意，按照合并的堆数循环 for(int i=1;i<=n-l+1;i++){//小心越界 int j=l+i-1;//j通过i,l得到 if(i!=j)f[i][j]=INT_MAX;//设为最大（初始化） for(int k=i;k<j;k++){//枚举中间点 f[i][j]=min(f[i][k]+f[k+1][j]+sum[j]-sum[i-1],f[i][j]);//转移方程 } } } printf("%d\n",f[1][n]);//输出 return 0; }-
CoderCharlie
@ 2023-5-28 13:47:58
?!
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0
世界上最难的题目
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize("Ofast") #pragma GCC optimize("inline") #pragma GCC optimize("-fgcse") #pragma GCC optimize("-fgcse-lm") #pragma GCC optimize("-fipa-sra") #pragma GCC optimize("-ftree-pre") #pragma GCC optimize("-ftree-vrp") #pragma GCC optimize("-fpeephole2") #pragma GCC optimize("-ffast-math") #pragma GCC optimize("-fsched-spec") #pragma GCC optimize("unroll-loops") #pragma GCC optimize("-falign-jumps") #pragma GCC optimize("-falign-loops") #pragma GCC optimize("-falign-labels") #pragma GCC optimize("-fdevirtualize") #pragma GCC optimize("-fcaller-saves") #pragma GCC optimize("-fcrossjumping") #pragma GCC optimize("-fthread-jumps") #pragma GCC optimize("-funroll-loops") #pragma GCC optimize("-fwhole-program") #pragma GCC optimize("-freorder-blocks") #pragma GCC optimize("-fschedule-insns") #pragma GCC optimize("inline-functions") #pragma GCC optimize("-ftree-tail-merge") #pragma GCC optimize("-fschedule-insns2") #pragma GCC optimize("-fstrict-aliasing") #pragma GCC optimize("-fstrict-overflow") #pragma GCC optimize("-falign-functions") #pragma GCC optimize("-fcse-skip-blocks") #pragma GCC optimize("-fcse-follow-jumps") #pragma GCC optimize("-fsched-interblock") #pragma GCC optimize("-fpartial-inlining") #pragma GCC optimize("no-stack-protector") #pragma GCC optimize("-freorder-functions") #pragma GCC optimize("-findirect-inlining") #pragma GCC optimize("-fhoist-adjacent-loads") #pragma GCC optimize("-frerun-cse-after-loop") #pragma GCC optimize("inline-small-functions") #pragma GCC optimize("-finline-small-functions") #pragma GCC optimize("-ftree-switch-conversion") #pragma GCC optimize("-foptimize-sibling-calls") #pragma GCC optimize("-fexpensive-optimizations") #pragma GCC optimize("-funsafe-loop-optimizations") #pragma GCC optimize("inline-functions-called-once") #pragma GCC optimize("-fdelete-null-pointer-checks") #include<iostream> #include<cstring> #include<cstdio> #include<cstring> using namespace std; struct node { int data,rev,sum; node *son[2],*pre; bool judge(); bool isroot(); void pushdown(); void update(); void setson(node *child,int lr); }lct[233]; int top,a,b; node *getnew(int x) { node *now=lct+ ++top; now->data=x; now->pre=now->son[1]=now->son[0]=lct; now->sum=0; now->rev=0; return now; } bool node::judge(){return pre->son[1]==this;} bool node::isroot() { if(pre==lct)return true; return !(pre->son[1]==this||pre->son[0]==this); } void node::pushdown() { if(this==lct||!rev)return; swap(son[0],son[1]); son[0]->rev^=1; son[1]->rev^=1; rev=0; } void node::update(){sum=son[1]->sum+son[0]->sum+data;} void node::setson(node *child,int lr) { this->pushdown(); child->pre=this; son[lr]=child; this->update(); } void rotate(node *now) { node *father=now->pre,*grandfa=father->pre; if(!father->isroot()) grandfa->pushdown(); father->pushdown();now->pushdown(); int lr=now->judge(); father->setson(now->son[lr^1],lr); if(father->isroot()) now->pre=grandfa; else grandfa->setson(now,father->judge()); now->setson(father,lr^1); father->update();now->update(); if(grandfa!=lct) grandfa->update(); } void splay(node *now) { if(now->isroot())return; for(;!now->isroot();rotate(now)) if(!now->pre->isroot()) now->judge()==now->pre->judge()?rotate(now->pre):rotate(now); } node *access(node *now) { node *last=lct; for(;now!=lct;last=now,now=now->pre) { splay(now); now->setson(last,1); } return last; } void changeroot(node *now) { access(now)->rev^=1; splay(now); } void connect(node *x,node *y) { changeroot(x); x->pre=y; access(x); } void cut(node *x,node *y) { changeroot(x); access(y); splay(x); x->pushdown(); x->son[1]=y->pre=lct; x->update(); } int query(node *x,node *y) { changeroot(x); node *now=access(y); return now->sum; } int main() { scanf("%d%d",&a,&b); node *A=getnew(a); node *B=getnew(b); //连边 Link connect(A,B); //断边 Cut cut(A,B); //再连边orz Link again connect(A,B); printf("%d\n",query(A,B)); return 0; } -
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CZOJ @ 2023-3-9 19:29:59
本题代码：
C
#include<stdio.h> int main() { int a,b; scanf("%d%d",&a,&b); printf("%d",a+b); return 0; }C++
#include<bits/stdc++.h> using namespace std; int main() { int a,b; cin>>a>>b; cout<<a+b; return 0; }Python
a,b=map(int,input().split()) print(a+b)
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