【实践课内】6.循环结构3

【实践课内】6.循环结构3/7-1 求简单交错序列前N项和.c

@@ -0,0 +1,20 @@
+#include <stdio.h>
+
+int main() {
+    int n;
+    double ans = 0;
+
+    scanf("%d", &n);
+
+    for (int i = 0; i < n; i++) {
+        if (i % 2 == 0) {
+            ans += 1.0 / (1 + 3 * i);
+        } else {
+            ans -= 1.0 / (1 + 3 * i);
+        }
+    }
+
+    printf("sum = %.3lf\n", ans);
+
+    return 0;
+}

【实践课内】6.循环结构3/7-2 21循环-求和2.c

@@ -0,0 +1,22 @@
+#include <math.h>
+#include <stdio.h>
+
+int main() {
+    long long n, ans = 0;
+
+    scanf("%lld", &n);
+
+    for (long long i = 1; i <= (long long)sqrt(n); i++) {
+        if (n % i == 0) {
+            ans += i;
+
+            if (i != n / i) {
+                ans += n / i;
+            }
+        }
+    }
+
+    printf("%lld\n", ans);
+
+    return 0;
+}

【实践课内】6.循环结构3/7-3 个位数.c

@@ -0,0 +1,22 @@
+#include <stdio.h>
+
+int main() {
+    int n;
+
+    scanf("%d", &n);
+
+    while (n >= 10) {
+        int s = 0;
+
+        while (n) {
+            s += n % 10;
+            n /= 10;
+        }
+
+        n = s;
+    }
+
+    printf("%d\n", n);
+
+    return 0;
+}

【实践课内】6.循环结构3/7-4 循环-分数矩阵.c

@@ -0,0 +1,34 @@
+#include <math.h>
+#include <stdio.h>
+
+int main() {
+    int n;
+
+    while (scanf("%d", &n) != EOF, n != 0) {
+        double ans = n;
+
+        for (int i = 2; i <= n; i++) {
+            ans += (double)(n * 2 - 2 * (i - 1)) / (double)i;
+        }
+
+        /*
+        for (int i = 1; i <= n; i++) {
+            // 对角线左侧
+            for (int j = i; j >= 2; j--) {
+                ans += 1.0 / j;
+            }
+
+            ans += 1;
+
+            // 对角线右侧
+            for (int j = 2; j <= n - i + 1; j++) {
+                ans += 1.0 / j;
+            }
+        }
+        */
+
+        printf("%.2lf\n", ans);
+    }
+
+    return 0;
+}