A+B没过，谁来救我一下

setpresetion @ 2024-01-29 14:21:26

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,i,j,k,maxn=0,f[1001000],ok[1000100];
struct node{
    int s,t,w;
}a[1001000];

bool valid(int x){
    memset(f,0,sizeof(f));
    for(i=1;i<=x;++i){
        f[a[i].s]+=a[i].w;
        f[a[i].t+1]-=a[i].w;
    }
    j=0;
    for(i=1;i<=n;++i){
        j+=f[i];
        if(j>ok[i])return false;
    }
    return true;
}
int main(){
    memset(f,0,sizeof(f));
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;++i)scanf("%d",&ok[i]);
    for(i=1;i<=m;++i)scanf("%d%d%d",&a[i].w,&a[i].s,&a[i].t);
    if(valid(m)){printf("0\n");return 0;}
    int l=1,r=m;
    while(l<r){
        int mid=(l+r)/2;
        if(valid(mid)) l=mid+1;
        else r=mid;
    }
    printf("-1\n%d\n",l);
    return 0;
}

by RaymondOccam @ 2024-02-17 15:47:10

@rickyxue 用printf干嘛？

把sync关了，跑cin和cout它不香吗

ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
// PS:不需要附加其他头文件

不满意？

火车头直接起飞

#include <iostream>
#pragma GCC optimize("Ofast", "inline", "-ffast-math")
#pragma GCC target("avx,sse2,sse3,sse4,mmx")
// PS:有些oj的测评机特别友(dú)善(liú),打比赛最好别用火车头优化了

by test_daily @ 2024-02-17 18:33:35

你干嘛要高精度，你直接long long就可以了

by qby1166 @ 2024-02-18 17:45:34

nb

by liuguanyu @ 2024-02-19 15:42:49

@rickyxue 6

by quxiangyu @ 2024-02-19 21:24:56

高精度是这样的：

#include<bits/stdc++.h>
using namespace std;
const int N = 1005;
struct bign
{
    int len,s[N];
    bign()  {  memset(s,0,sizeof(s));  len=1;  }
    bign(int num)  {  *this=num; }
    bign(char *num) { *this=num; }
    bign operator =(int num)
    {
        char c[N];
        sprintf(c,"%d",num);
        *this=c;
        return *this;
    }
    bign operator =(const char *num)
    {
        len=strlen(num);
        for (int i=0;i<len;i++) s[i]=num[len-1-i]-'0';
        return *this;
    }
    string str()
    {
        string res="";
        for (int i=0;i<len;i++) res=(char)(s[i]+'0')+res;
        return res;
    }
    void clean()
    {
        while (len>1&&!s[len-1]) len--;
    }
    bign operator +(const bign &b)
    {
        bign c;    
        c.len=0;
        for (int i=0,g=0;g||i<len||i<b.len;i++)
        {
            int x=g;
            if (i<len) x+=s[i];
            if (i<b.len) x+=b.s[i];
            c.s[c.len++]=x%10;
            g=x/10;
        }
        return c;
    }
    bign operator -(const bign &b)
    {
        bign c;
        c.len=0;
        int x;     
        for (int i=0,g=0;i<len;i++)
        {
            x=s[i]-g;
            if (i<b.len) x-=b.s[i];
            if (x>=0) g=0;
            else{          
                x+=10;
                g=1;
            };
            c.s[c.len++]=x;
        }
        c.clean();
        return c;
    }
    bign operator *(const bign &b)
    {
        bign c;
        c.len=len+b.len;
        for (int i=0;i<len;i++) for (int j=0;j<b.len;j++) c.s[i+j]+=s[i]*b.s[j];
        for (int i=0;i<c.len-1;i++) { c.s[i+1]+=c.s[i]/10; c.s[i]%=10; }
        c.clean();
        return c;  
    }
    bool operator <(const bign &b)
    {
        if (len!=b.len) return len<b.len;
        for (int i=len-1;i>=0;i--)
             if (s[i]!=b.s[i]) return s[i]<b.s[i];
        return false;
    }
    bign operator +=(const bign &b)
    {
        *this=*this+b;
        return *this;
    }
    bign operator -=(const bign &b)
    {
        *this=*this-b;
        return *this;
    }  
};
istream& operator >>(istream &in,bign &x)
{
  string s;
  in>>s;
  x=s.c_str();
  return in;
}
ostream& operator <<(ostream &out,bign &x)
{
    out<<x.str();
    return out;
}
int main(){
    bign a,b,c;
    ios::sync_with_stdio(false);
    cin>>a>>b;
//    cout<<a<<endl;
//    cout<<b<<endl;
    c=a+b;
    cout<<c<<endl;
    return 0;
}

by quxiangyu @ 2024-02-19 21:27:24

但是这题不能过

by RaymondOccam @ 2024-02-21 20:54:13

@xiaoshumiao 是你不懂吧

by xiaoshumiao @ 2024-02-21 21:36:39

@mini123 你有本事用 int 过一遍试试

by zhengkaiyang @ 2024-02-25 11:02:35

6

by RaymondOccam @ 2024-02-29 21:00:25

@xiaoshumiao

本来就不是int的问题好不好？