题目链接如下:
这道题要求“Follow the output for each region with a blank line.”,而不是region之间加空行,我在这个点上卡了很久……
然后我有点把题复杂化了,其实不用算currArea那么复杂,直接用一个for循环,循环到哪就有几块地被淹了,挺简单的....下面这个代码写得比较好。
https://www.cnblogs.com/zyb993963526/p/6297474.html
我的(复杂)代码如下:
- #include
- #include
- #include
- // #define debug
-
- int m, n, tot, currArea, pivot, kase = 0;
- double waterLevel;
-
- int main(){
- #ifdef debug
- freopen("0.txt", "r", stdin);
- freopen("1.txt", "w", stdout);
- #endif
- while(scanf("%d %d", &m, &n) == 2 && m && n){
- std::vector<int> vec(m * n);
- for(int i = 0; i < m * n; ++i){
- scanf("%d", &vec[i]);
- }
- sort(vec.begin(), vec.end());
- scanf("%d", &tot);
- currArea = 0;
- pivot = -1;
- waterLevel = vec[0];
- while(tot){
- do{
- pivot++;
- currArea += 100;
- } while(pivot < vec.size() - 1 && vec[pivot + 1] == vec[pivot]);
- if(pivot == vec.size() - 1 || tot * 1.0 / currArea <= vec[pivot + 1] - vec[pivot]){
- waterLevel = vec[pivot] + tot * 1.0 / currArea;
- break;
- }
- tot -= currArea * (vec[pivot + 1] - vec[pivot]);
- }
- printf("Region %d\nWater level is %.2f meters.\n%.2f percent of the region is under water.\n\n", ++kase, waterLevel, (pivot + 1) * 100.0 / m / n);
- }
- #ifdef debug
- fclose(stdin);
- fclose(stdout);
- #endif
- return 0;
- }