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Update: 0416 使用STL求和

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Original file line number	Diff line number	Diff line change
`@@ -143,15 +143,12 @@ dp[j]的数值一定是小于等于j的。`
`143`	`143`	`class Solution {`
`144`	`144`	`public:`
`145`	`145`	`bool canPartition(vector<int>& nums) {`
`146`		`- int sum = 0;`
`147`		`-`
	`146`	`+ // 使用标准库函数便捷求和`
	`147`	`+ int sum = accumulate(nums.begin(), nums.end(), 0);`
`148`	`148`	`// dp[i]中的i表示背包内总和`
`149`	`149`	`// 题目中说:每个数组中的元素不会超过 100,数组的大小不会超过 200`
`150`	`150`	`// 总和不会大于20000,背包最大只需要其中一半,所以10001大小就可以了`
`151`	`151`	`vector<int> dp(10001, 0);`
`152`		`- for (int i = 0; i < nums.size(); i++) {`
`153`		`- sum += nums[i];`
`154`		`- }`
`155`	`152`	`if (sum % 2 == 1) return false;`
`156`	`153`	`int target = sum / 2;`
`157`	`154`

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