无效的下一个尺寸和两次免费展示错误

Question

我这里有一些非常简单的代码行，但是它们给了我一些严重的错误消息。

1。我的代码：

    #include <bits/stdc++.h>

    using namespace std;

    void solve( int n, unsigned long long k, int x){
        unsigned long long divi = 1000000007;
        std::vector< std::vector < unsigned long long > > dpArray( n, std::vector < unsigned long long >( 3, 0));
        dpArray[0][0] = 1;
        for(int index = 1; index < n; index++){
            dpArray[0][index] = (dpArray[1][index - 1] + dpArray[2][index - 1])%divi;
            dpArray[1][index] = (dpArray[0][index - 1] + dpArray[2][index - 1])%divi;
            dpArray[2][index] = ((k - 2)*dpArray[1][index])%divi;
        }

        int ans = 0;
        if(x == 1){
            ans = (int)dpArray[0][n - 1];
        }
        else{
            ans = (int)dpArray[1][n - 1];
        }
        std::cout << ans << std::endl;
    }

    int main(){
        int n = 0, x = 0;
        unsigned long long k;
        std::cin >> n >> k >> x;
        solve( n, k, x);
        return 0;
    }

1. 下面是我输入100000 10000 1

   时的错误消息

   151337967
   Error in `./a.out': free(): invalid next size (fast): 0x0000000000818050
   ======= Backtrace: =========
   /lib/x86_64-linux-gnu/libc.so.6(+0x777e5)[0x7f90f9af97e5]
   /lib/x86_64-linux-gnu/libc.so.6(+0x8037a)[0x7f90f9b0237a]
   /lib/x86_64-linux-gnu/libc.so.6(cfree+0x4c)[0x7f90f9b0653c]
   ./a.out[0x401aee]
   ./a.out[0x401961]
   ./a.out[0x401730]
   ./a.out[0x401422]
   ./a.out[0x401210]
   ./a.out[0x401c77]
   ./a.out[0x401ab4]
   ./a.out[0x40190d]
   ./a.out[0x4016b1]
   ./a.out[0x40130c]
   ./a.out[0x400f86]
   ./a.out[0x401078]
   /lib/x86_64-linux-gnu/libc.so.6(__libc_start_main+0xf0)[0x7f90f9aa2830]
   ./a.out[0x400bb9]
   

stdout中的第一行是正确的答案，但随后出现此错误！

Answer 1

std::vector< std::vector < unsigned long long > > dpArray( n, std::vector < unsigned long long >( 3, 0))

这将创建大小为[n][3]的向量。因此您将在以下代码中访问索引

for(int index = 1; index < n; index++){
        dpArray[0][index] = (dpArray[1][index - 1] + dpArray[2][index - 1])%divi;
        dpArray[1][index] = (dpArray[0][index - 1] + dpArray[2][index - 1])%divi;
        dpArray[2][index] = ((k - 2)*dpArray[1][index])%divi;
    }