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GitHub - jzplp/aoapc-UVA-Answer: 算法竞赛入门经典 例题和习题答案 刘汝佳 第二版

根据书上的方法来做，是比较简单的题目。关键在于知道等体积时的枚举法。不过数据大小可能很大，虽然输入可以用int处理，但是 体积*价值 后，需要long long才存的下。

#include<stdio.h>int n, s1, v1, s2, v2;long long getCount(int num, int type) {long long sumV = 0, maxV = 0;for(int i = 0; i <= num; ++i) {if(type == 1) {if((n - i * s1) < 0) continue;sumV = (long long)i * v1 + ((long long)(n - i * s1) / s2) * v2;} else {if((n - i * s2) < 0) continue;sumV = (long long)i * v2 + ((long long)(n - i * s2) / s1) * v1;}// printf("%lld %lld\n", sumV, maxV);if(sumV > maxV) maxV = sumV;}return maxV;
}long long getRes() {int a = n / s1;int b = n / s2;int type;if ((long long)s2 * v1 > (long long)s1 * v2) {type = 2; } else if((long long)s2 * v1 < (long long)s1 * v2) {type = 1;} else {if(s1 > s2) {type = 1;} else {type = 2; } }int typeNum = type == 1 ? s2 : s1;// printf("%d %d %d %d\n", a, b, type, typeNum);if(a <= b && a <= typeNum) {return getCount(a, 1);}else if(b <= a && b <= typeNum) {return getCount(b, 2);}else if(typeNum <= a && typeNum <= b) {return getCount(typeNum, type);}return getCount(a, 1);
}int main() {int t, ti = 0;scanf("%d", &t);while(ti++ != t) {scanf("%d %d %d %d %d", &n, &s1, &v1, &s2, &v2);// printf("%d\n", n);printf("Case #%d: %lld\n", ti, getRes());}return 0; 
}