C++ 一个字符串只包含K和D。K表示杀人，D表示死亡。计算连续杀敌次数，阵亡则归零。

题：

 

首先笔者的第一思路是前后两个字符串比较。如果相同index++。

#include<iostream>
#include<stdio.h>
#include<string>
using namespace std;
int main()
{
	string s;
	cout << "6请输入字符串：" << endl;
	cin >> s;
	string name = "";
	unsigned int iSize = s.size();
	int currVal = 0;
	for (int i = 0; i <iSize; i++)
	{
		int j = i;
		if (s[j] == s[j+1])
			currVal++;
		else
			currVal = 0;
	}
	cout << currVal << endl;
	switch (currVal)
	{
	case 0:name = "a Foolish Man"; break;
	case 1:name = "a Foolish Man"; break;
	case 2:name = "a Foolish Man"; break;
	case 3:name = "a Foolish Man"; break;
	case 4:name = "Dominating"; break;
	case 5:name = "a Mega-Kill"; break;
	case 6:name = "Unstoppable"; break;
	case 7:name = "Wicked Sick"; break;
	case 8:name = "a M-m-m-m...Monster Kill"; break;
	case 9:name = "Godlike"; break;
	default:
		name = "Beyond Godlike";
		break;
	}
	std::cout << name;
	system("pause");
	return 0;
}

当我输入题示要求后ok。但是当输入DDDKKKK是期待输出是4但是输出为0.

分析结果是因为在j和j+1比较时j+1会出界。所以最后会置为0

后来网上搜索字符串中字符连续个数：

#include"string"
#include"iostream"
using namespace std;
int  main()
{
    string str;
    cin>>str;
    int lenght=str.length();
    int index=0;
    int i;
    int j;
    for(i=0;i<length;i=j)
    {
        index=1;
        for(j=i+1;j<length;++j)
        {
            if(str[i]==str[j])
                index++;
            else 
                break;
        }
        cout<<str[i]<<index;
    }
    cout<<endl;
    system("pause");
    return 0;
}

这个程序在数连续字符时完全没有问题。它采用起始为1，然后和后续比较，若不相同直接从不相同处开始比较的思想。但是如果最后是一个D或者多个D时我需要输出为0.为一个K时输出为1.。。试了多次依然没有成功

最后。突然用自己之前一个错误s[+1]起始就一直是s[0]的错误解决，就是每个字符都和K进行比较

#include<iostream>
#include<stdio.h>
#include<string>
using namespace std;
int main()
{
	string s ;
	cout << "3请输入字符串：" << endl;
	cin >> s;
	string name="";
	unsigned int iSize = s.size();
	int currVal = 0;
	for (int i = 0; i <iSize; i++)
	{
		if (s[i]=='K')
			currVal++;
		else
			currVal = 0;
	}
	cout << currVal << endl;
	switch (currVal)
	{
	case 0:name = "a Foolish Man"; break;
	case 1:name = "a Foolish Man"; break;
	case 2:name = "a Foolish Man"; break;
	case 3:name = "a Foolish Man"; break;
	case 4:name = "Dominating"; break;
	case 5:name = "a Mega-Kill"; break;
	case 6:name = "Unstoppable"; break;
	case 7:name = "Wicked Sick"; break;
	case 8:name = "a M-m-m-m...Monster Kill"; break;
	case 9:name = "Godlike"; break;
	default:
		name = "Beyond Godlike";
		break;
	}
	std::cout<<name;
	system("pause");
	return 0;
}

此时满足题意。

参考自https://blog.youkuaiyun.com/qq_27302361/article/details/51111537