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添加 1005.K次取反后最大化的数组和.md C版本

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`@@ -211,5 +211,44 @@ var largestSumAfterKNegations = function(nums, k) {`
`211`	`211`	`};`
`212`	`212`	```
`213`	`213`
	`214`	`+### C`
	`215`	+```c
	`216`	`+#define abs(a) (((a) > 0) ? (a) : (-(a)))`
	`217`	`+`
	`218`	`+// 对数组求和`
	`219`	`+int sum(int *nums, int numsSize) {`
	`220`	`+ int sum = 0;`
	`221`	`+`
	`222`	`+ int i;`
	`223`	`+ for(i = 0; i < numsSize; ++i) {`
	`224`	`+ sum += nums[i];`
	`225`	`+ }`
	`226`	`+ return sum;`
	`227`	`+}`
	`228`	`+`
	`229`	`+int cmp(const void* v1, const void* v2) {`
	`230`	`+ return abs((int)v2) - abs((int)v1);`
	`231`	`+}`
	`232`	`+`
	`233`	`+int largestSumAfterKNegations(int* nums, int numsSize, int k){`
	`234`	`+ qsort(nums, numsSize, sizeof(int), cmp);`
	`235`	`+`
	`236`	`+ int i;`
	`237`	`+ for(i = 0; i < numsSize; ++i) {`
	`238`	`+ // 遍历数组,若当前元素<0则将当前元素转变,k--`
	`239`	`+ if(nums[i] < 0 && k > 0) {`
	`240`	`+ nums[i] *= -1;`
	`241`	`+ --k;`
	`242`	`+ }`
	`243`	`+ }`
	`244`	`+`
	`245`	`+ // 若遍历完数组后k还有剩余(此时所有元素应均为正),则将绝对值最小的元素nums[numsSize - 1]变为负`
	`246`	`+ if(k % 2 == 1)`
	`247`	`+ nums[numsSize - 1] *= -1;`
	`248`	`+`
	`249`	`+ return sum(nums, numsSize);`
	`250`	`+}`
	`251`	+```
	`252`	`+`
`214`	`253`	`-----------------------`
`215`	`254`	`<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>`

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