【算法刷题】二分法（2）

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A+B and C

Given three integers A, B and C in [-263, 263), you are supposed to tell whether A+B > C.

#include <cstdio>
#include <algorithm>
 
using namespace std;
 
int sum[1005];
int main() {
int n;
    scanf("%d",&n);
    for (int i = 1; i <= n; ++i) {
        long long a,b,c;
        scanf("%lld%lld%lld",&a,&b,&c);
        long long ab = a+b;
        if(ab>c) sum[i] = 1;
        else sum[i] = 0;
        if(a>0 && b>0 && ab<0) sum[i] = 1;
        if(a<0 && b<0 && ab>=0) sum[i] = 0;
    }
    for (int i = 1;i <= n;++i) {
        if(sum[i]) printf("Case #%d: true\n",i);
        else printf("Case #%d: false\n",i);
    }
    return 0;
}

PS:

• long long 的范围时[-2^63^,2^63^)，因此两个long long类型的整数相加可能会溢出，需要通过A,B,A+B的符号来判断是否溢出
• long long 存储2^63^的时候会自动变成-2^63^，无法区分左右边界。

A+B in Hogwarts

If you are a fan of Harry Potter, you would know the world of magic has its own currency system -- as Hagrid explained it to Harry, "Seventeen silver Sickles to a Galleon and twenty-nine Knuts to a Sickle, it's easy enough." Your job is to write a program to compute A+B where A and B are given in the standard form of "Galleon.Sickle.Knut" (Galleon is an integer in [0, 107], Sickle is an integer in [0, 17), and Knut is an integer in [0, 29)).

#include <cstdio>
#include <algorithm>
 
using namespace std;
 

int main() {
    int A=0,B=0,C=0;
    scanf("%d.%d.%d",&A,&B,&C);
    int a=0,b=0,c=0;
    scanf("%d.%d.%d",&a,&b,&c);
    A+=a;
    B+=b;
    C+=c;
    
    B+=(C/29);
    C%=29;
    A+=(B/17);
    B%=17;
    printf("%d.%d.%d",A,B,C);
    return 0;
}

Ps:审题要清晰。

Kuchiguse

The Japanese language is notorious for its sentence ending particles. Personal preference of such particles can be considered as a reflection of the speaker's personality. Such a preference is called "Kuchiguse" and is often exaggerated artistically in Anime and Manga. For example, the artificial sentence ending particle "nyan~" is often used as a stereotype for characters with a cat-like personality: Itai nyan~ (It hurts, nyan~) Ninjin wa iyada nyan~ (I hate carrots, nyan~) Now given a few lines spoken by the same character, can you find her Kuchiguse?

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
 
using namespace std;

string rev(string str){
    int len = str.length();
    for (int i = 0; i < len/2;++i){
        char temp = str[i];
        str[i] = str[len-i-1];
        str[len-i-1] = temp;
    }
    return str;
}
string abc[259];

int main() {
    int n;

    scanf("%d",&n);
    getchar();
    for (int i = 1; i <= n; ++i) {
        getline(cin,abc[i]);
        abc[i] = rev(abc[i]);
    }
    int short1;
    bool a=1;
    string res="";
    for (int i = 1; i < n; ++i) {
        if(abc[i].length()<abc[i+1].length()) short1=abc[i].length();
    }
    for (int i = 0; i< short1;++i){
        char temp;
        for(int j = 1; j < n&& a==1;++j){
            temp = abc[j][i];
            if(abc[j][i]==abc[j+1][i]) continue;
            else a=0;
        }
        if(a==1) res+=temp;
        else break;
    }
    res = rev(res);
    if(res == "")
        printf("nai");
    else
    for (int i = 0; i < res.length();++i){
            printf("%c",res[i]);
    }
    return 0;
}

思考和总结

对于二分法来说，最麻烦的可能就是边界的判断了，记得有一次我在比赛的时候，我一直卡在一道题上面，不断提交导致罚时，以至于最后痛失名次。 对于边界的判断，一定要提前想好，算好，测试的时候选择一些边界值来计算。

PS: 可以用algorithm的reverse函数来反转字符串。