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HDU 5982.Relic Discovery(2016 CCPC 青岛 A)|
2016-12-10 · via OhYee 博客

这是一篇最后编辑于 8 年前 的文章,其内容可能与目前实际情况差异较大,请注意甄别

题目

{% raw %}

{% endraw %}

Recently, paleoanthropologists have found historical remains on an island in the Atlantic Ocean. The most inspiring thing is that they excavated in a magnificent cave and found that it was a huge tomb. Inside the construction,researchers identified a large number of skeletons, and funeral objects including stone axe, livestock bones and murals. Now, all items have been sorted, and they can be divided into N types. After they were checked attentively, you are told that there are Ai items of the i-th type. Further more, each item of the i-th type requires Bi million dollars for transportation, analysis, and preservation averagely. As your job, you need to calculate the total expenditure.

{% raw %}


{% endraw %}

The first line of input contains an integer T which is the number of test cases. For each test case, the first line contains an integer N which is the number of types. In the next N lines, the i-th line contains two numbers Ai and Bi as described above. All numbers are positive integers and less than 101.

{% raw %}


{% endraw %}

For each case, output one integer, the total expenditure in million dollars.

{% raw %}




{% endraw %}
1
2
1 2
3 4
{% raw %}


{% endraw %}
14
{% raw %}




{% endraw %}

题解

看样例基本就是对于 T 组数据
每组数据有 nAB
求这 nA * B

读题验证思路无误

代码

```cpp Relic Discovery https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份 /** #define debug #include

//*/

#include
#include
#include
using namespace std;

int main() {
#ifdef debug
freopen("in.txt", "r", stdin);
int START = clock();
#endif

int T;
cin >> T;
while (T--) {
    int n, sum = 0;
    cin >> n;
    while (n--) {
        int A, B;
        cin >> A >> B;
        sum += A * B;
    }
    cout << sum << endl;
}

#ifdef debug
printf("Time:%.3f s.\n", double(clock() - START) / CLOCKS_PER_SEC);
#endif
return 0;
}

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