class Solution {

public:

static bool cmp(int a, int b) {

return a >= b;

}

int breakfastNumber(vector<int>& staple, vector<int>& drinks, int x) {

int ans = 0;

int stapleLen = staple.size();

cout << stapleLen << endl;

int drinksLen = drinks.size();

cout << drinksLen << endl;

sort(staple.begin(), staple.end(), cmp);

sort(drinks.begin(), drinks.end());

int j = 0;

for (int i = 0; i < stapleLen; i++) {

ans += j;

ans %= 1000000007;

for (; j < drinksLen; j++) {

if (staple[i] + drinks[j] <= x) {

ans++;

ans %= 1000000007;

} else {

break;

}

}

}

return ans;

}

};

提示信息：

=================================================================

==42==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x7f22768b87fc at pc 0x00000034d6ad bp 0x7ffc746101b0 sp 0x7ffc746101a8

READ of size 4 at 0x7f22768b87fc thread T0

#6 0x7f2279d720b2 (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)

0x7f22768b87fc is located 4 bytes to the left of 524288-byte region [0x7f22768b8800,0x7f2276938800)

allocated by thread T0 here:

#6 0x7f2279d720b2 (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)

Shadow bytes around the buggy address:

0x0fe4ced0f0a0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa

0x0fe4ced0f0b0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa

0x0fe4ced0f0c0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa

0x0fe4ced0f0d0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa

0x0fe4ced0f0e0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa

=>0x0fe4ced0f0f0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa[fa]

0x0fe4ced0f100: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

0x0fe4ced0f110: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

0x0fe4ced0f120: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

0x0fe4ced0f130: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

0x0fe4ced0f140: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

Shadow byte legend (one shadow byte represents 8 application bytes):

Addressable: 00

Partially addressable: 01 02 03 04 05 06 07

Heap left redzone: fa

Freed heap region: fd

Stack left redzone: f1

Stack mid redzone: f2

Stack right redzone: f3

Stack after return: f5

Stack use after scope: f8

Global redzone: f9

Global init order: f6

Poisoned by user: f7

Container overflow: fc

Array cookie: ac

Intra object redzone: bb

ASan internal: fe

Left alloca redzone: ca

Right alloca redzone: cb

Shadow gap: cc

==42==ABORTING

类解决方案 { 公共: 静态bool cmp(int a，int b) { 返回a >= b; } int breakfastNumber(vector<int>& stub，vector<int>& 饮料，int x) { int = 0; int stapleLen = 订书钉.尺寸 (); cout <stapleLen <endl; int drinksLen = 饮料尺寸 (); cout << drinksLen <endl; sort (studin.begin()，stapin.end()，cmp); 排序 (饮料.开始 () 、饮料.结束 ()); int j = 0; for (int i = 0; i < stapleLen; i ++) { ans + = j; ans % = 1000000007; 用于 (; j < drinksLen; j ++) { 如果 (主食 [i] + 饮料 [j] <= x) { ans ++; ans % = 1000000007; } else { 打破; } } } 退货; } }; 提示信息: ================================================================= = = 42 = ERROR: AddressSanitizer: 堆-缓冲区-在pc 0x00000034d6ad bp处的地址0x7f22768b87fc 746101b0 sp 0x7ffc746101a8 在0x7f22768b87fc线程T0读取大小为4 #6 0x7f2279d720b2 (/lib/x86_64-linux-gnu/libc.so.6 + 0x270b2) 0x7f22768b87fc位于524288字节区域左侧4字节 [0x7f22768b8800，0x7f2276938800) 由线程T0在这里分配: #6 0x7f2279d720b2 (/lib/x86_64-linux-gnu/libc.so.6 + 0x270b2) 围绕越野车地址的阴影字节: 0x0fe4ced0f0a0: fa 0x0fe4ced0f0b0: fa 0x0fe4ced0f0c0: fa 0x0fe4ced0f0d0: fa 0x0fe4ced0f0e0: fa =>0x0fe4ced0f0f0: fa[fa] 0x0fe4ced0f100: 00 0x0fe4ced0f110: 00 0x0fe4ced0f120: 00 0x0fe4ced0f130: 00 0x0fe4ced0f140: 00 阴影字节图例 (一个阴影字节代表8个应用程序字节): 可寻址: 00 部分可寻址: 01 02 03 04 05 06 07 堆左redzone: fa 自由堆区域: fd 堆栈左redzone: f1 堆栈中redzone: f2 堆栈右红区: f3 返回后堆栈: f5 作用域后堆栈使用: f8 全球redzone: f9 全球init订单: f6 被用户中毒: f7 容器溢出: fc 数组cookie: ac Intra对象redzone: bb ASan内部: fe 左alloca redzone: ca 右alloca redzone: cb 阴影间隙: cc = = 42 = = 引产

请问各位大佬，怎么在MATLAB中怎么计算2个经纬度之间的距离呀？

也许你在飞

2个回复