二分 50分求助

somek_ @ 2023-10-12 19:06:55

#include <bits/stdc++.h>
using namespace std;

int main() {
    int a, b;
    cin >> a >> b;
    long long l = 1, r = pow(2, 31) - 2;
    while (l < r) {
        long long mid = (l + r) / 2;
        if (a + b <= mid)
            r = mid;
        else
            l = mid + 1;
    }
    cout << r;
}

by zhanghe724_1 @ 2023-10-20 19:38:00

搞错了，再来

#include<bits/stdc++.h>
using namespace std;
inline __int128 read(){
    __int128 x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
inline void write(__int128 x){
    if(x<0){
        putchar('-');
        x=-x;
    }
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
int main(){
    __int128 a=read();
    __int128 b=read();
    write(a+b);
    return 0;
}

by zhanghe724_1 @ 2023-10-20 19:43:15

@lijingshu_304775

我还真是刚知道这题用int能AC，以前都用long long/高精度做的

by xiao12chen @ 2023-11-17 21:47:01

多此一举

by zengyukai2012 @ 2024-01-04 21:31:29

：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int a, b;
    cin >> a >> b;
    long long l = -2e9, r = 2e9;
//这里用long long是因为若
//l=1000000000，r=2000000000时
//l+r会越界，导致程序运行时间变长！！！
    while (l < r)
    {
        int mid = (l + r) >> 1;
//不用/2是因为c++中除法是向零取整，
//对负数操作时会有问题，如：
//(-123)/2=-61
        if (a + b <= mid)
            r = mid;
        else
            l = mid + 1;
    }
    cout << r;
    return 0;//好习惯（虽然我一般不加）
}

by K1ondra @ 2024-02-18 11:09:57

高精没考虑负数的范围啊

by K1ondra @ 2024-02-18 11:23:20

#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream> 
using namespace std;

char ch[10010];
int a[10010],b[10010],la,lb;

int main() {
    cin>>ch+1;
    if(ch[1]=='-' ){
        la=strlen(ch+1)-1;
        for( int i=2;i<=la+1;++i)a[la+1-i+1]=-(ch[i]-'0');
    } else {
        la = strlen(ch+1);
        for( int i=1;i<=la;++i)a[la-i+1]=ch[i]-'0';
    }

    cin>>ch+1;
    if(ch[1]=='-'){
        lb=strlen(ch+1)-1;
        for(int i=2;i<=lb+1;++i)b[lb+1-i+1]=-(ch[i]-'0');
    } else {
        lb = strlen(ch+1);
        for( int i=1;i<=lb;++i)b[lb-i+1]=ch[i]-'0';
    }
    la=max(la,lb)+1;
    for(int i=1;i<la;++i)a[i]+=b[i];
    for(int i = 1; i<la;++i){ 
        a[i+1]+=a[i]/10;
        a[i]%=10;
        if(a[i]<0){
            a[i]+=10;
            --a[i+1];
        }
    }
    while(la>1&&a[la]==0)--la;
    if(a[la]<0){
        cout<<'-';
        for( int i=1;i<=la;++i)a[i]=-a[i];
        for( int i=1;i<la;++i){
            if(a[i]<0){
                a[i]+=10;
                --a[i+1];
            }
        }
        while(la>1&&a[la]==0)--la;
        for(int i=la;i>=1;--i)cout<<a[i];
    } else
    for(int i=la;i>=1;--i )cout<<a[i];
    cout<<endl;
    return 0;
}

你的高精度没有讨论负数的范围

by HUDSI1265 @ 2024-02-19 15:08:24

@somek_ 因为它相加可能是负数，或者超int，所以你要开longlong，左端点变成-2^63-1,右端点变成2^63-1